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Thread: Chapter 8: Further Applications of Integration

  1. #1 Chapter 8: Further Applications of Integration 
    Forum Bachelors Degree Demen Tolden's Avatar
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    Link to book index:
    http://www.thescienceforum.com/Calcu...ions-8994t.php

    Chapter 8: Further Applications of Integration

    So far things don't seem too different. In 8.1 I am introduced to the method used to measure the lengths of smooth curves.

    L = ∫ (1 + (f '(x) )<sup>2</sup> )<sup>1/2</sup> dx

    It seems though that it doesn't take much for integrals like these to be either very difficult or impossible to solve. Most versions of an (f '(x) )<sup>2</sup> come out to be three part polynomials. Finding the right way to work around the square root can get pretty difficult. So far I've spent a lot of time just staring at what I've wrote on the paper, trying to think my way through it.

    There is a problem I am working on right now where I have to choose to either try to find the integral of
    ∫ -sec Θ csc<sup>2</sup> Θ dΘ

    or
    ∫ sec<sup>2</sup> Θ csc Θ dΘ

    I've tried to work out both of these, but I seem to always get some kind of natural log that makes things more difficult.


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  3. #2  
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    Oh wow. I don't know what I was thinking.

    u = sec Θ
    dv = -csc<sup>2</sup> Θ
    du = sec Θ tan Θ
    v = cot Θ

    A = csc Θ - ∫ sec Θ dΘ
    A = csc Θ - ln (secΘ + tanΘ)


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  4. #3  
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    So are you in a Calculus class, or just reading the book for fun? It's neat to see somebody who is so enthused about the subject, because most people I have met do not seem to understand it or enjoy it.
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  5. #4  
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    The main reason I am learning calculus on my own because I feel that, since I am not a kid anymore, I have a responsibility to know why the world is the way it is, but also since I am going to be learning it anyway I can also use this to find myself a different type of job and get more variety out of life.

    I also love learning. There isn't really any subject I would not like to learn more about.
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  6. #5  
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    I don't quite understand 8.2 yet. What am I doing wrong on this problem?

    5) Find the area of the surface obtained by rotating the curve about the x axis.

    y = x<sup>3</sup> where 0 < x < 2 (This sign < is less than or equal.)

    My work:
    y = x<sup>3</sup>
    y' = 3x<sup>2</sup>
    (y')<sup>2</sup> = 9x<sup>4</sup>

    S = ∫ 2πx(1 + 9x<sup>4</sup>)<sup>1/2</sup> dx

    x = 3<sup>-1/2</sup>tan<sup>1/2</sup>Θ
    dx = 3<sup>-1/2</sup>2<sup>-1</sup>tan<sup>-1/2</sup>Θ
    tan Θ = 3x<sup>2</sup>

    S = ∫ 3<sup>-1</sup>π sec Θ
    S = 3<sup>-1</sup>π ln ( sec Θ + tan Θ )
    S = 3<sup>-1</sup>π ln ( (9x<sup>4</sup> + 1)<sup>1/2</sup> + 3x<sup>2</sup> )
    S = 3<sup>-1</sup>π( ln ( ( (3<sup>2</sup>2</sup>4</sup> + 1)<sup>1/2</sup> + 3<sup>1</sup>2<sup>2</sup>) / (2<sup>2</sup>) )

    I've been typically leaving everything in exponents since it seems easier to me to work with these problems in this way.

    book answer:
    π(145(145<sup>1/2</sup>) - 1)/27
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  7. #6  
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    I don't think you're getting the right expression for dx. What's the derivative of (tan Θ)<sup>1/2</sup>?
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  8. #7  
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    you are right.

    if
    x = tan<sup>1/2</sup> Θ

    then
    dx = (1/2) tan<sup>-1/2</sup> Θ sec<sup>2</sup> Θ
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  9. #8  
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    Is the Surface Area equation set up right though? That is really the one part of this problem that I am unsure about how to do. I've worked it out again and got it wrong.

    S = ∫<sub>0</sub><sup>2</sup> 2πx<sup>3</sup>(1 + 9x<sup>4</sup>)<sup>1/2</sup> dx

    This time I've worked it out to:
    S = ∫ 3<sup>-2</sup>π tan Θ sec<sup>3</sup>Θ dΘ
    S = 3<sup>-3</sup>π [ sec<sup>3</sup> Θ]<sub>0</sub><sup>2</sup>
    3<sup>-3</sup>π((13)<sup>3/2</sup> -1 )

    which of course is still wrong.
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  10. #9  
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    Yeah, excuse me, I should have noticed that the way you set it up the first time is wrong. The second time looks good to me, though. However, note that you're making things too hard on yourself: you can just substitute u = 9x<sup>4</sup>+1, du = 12x<sup>3</sup>dx. I didn't work it out, but I think what I get is different from you.
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  11. #10  
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    Let me just list off a few area equations to make sure I have this right. All the > signs and < signs include equals.

    #8 to #12: Find the area of the surface obtained by rotating the curve about the x axis.

    8 ) y = cos 2x
    0 < x < π/6

    My work:
    y' = -2 sin 2x
    (y')<sup>2</sup> = 4sin<sup>2</sup> 2x

    A = ∫<sub>0</sub><sup>π/6</sup> 2π( cos 2x)(1 + 4 sin<sup>2</sup> 2x )<sup>1/2</sup> dx

    9) y = cosh x
    0 < x < 1

    My work:
    y' = sinh x
    (y')<sup>2</sup> = sinh<sup>2</sup> x

    A = ∫<sub>0</sub><sup>1</sup> 2π(cosh x)(1 + sinh<sup>2</sup> x)<sup>1/2</sup> dx

    10) y = (x<sup>3</sup> / 6) + ( 1 / 2x )
    (1/2) < x < 1

    My work:
    y' = (x<sup>2</sup> / 2) - ( 1 / 2x<sup>2</sup> )
    (y')<sup>2</sup> = (x<sup>4</sup> / 4) - ( 1 / 4x<sup>4</sup> ) - (1/2)

    A = ∫<sub>(1/2)</sub><sup>1</sup> 2π( (x<sup>3</sup> / 6 ) + ( 1 / 2x ) )(1 + ( x<sup>4</sup> / 4 ) - ( 1 / 4x<sup>4</sup> ) - (1/2) )<sup>1/2</sup> dx

    11) x = (1/3)(y<sup>2</sup> + 2)<sup>3/2</sup>
    1 < y < 2

    My work:
    x' = y(y<sup>2</sup> + 2)<sup>1/2</sup>
    (x')<sup>2</sup> = y<sup>4</sup> + 2y<sup>2</sup>

    A = ∫<sub>1</sub><sup>2</sup> 2π(3<sup>-1</sup>( y<sup>2</sup> + 2)<sup>3/2</sup>)(1 + y<sup>4</sup> + 2y<sup>2</sup> )<sup>1/2</sup> dx

    12) x = 1 + 2y<sup>2</sup>
    1 < y < 2

    My work:
    x' = 4y
    (x')<sup>2</sup> = 16y<sup>2</sup>

    A = ∫<sub>1</sub><sup>2</sup>2π(1 + 2y<sup>2</sup>)(1 + 16y<sup>2</sup>)<sup>1/2</sup> dx

    Am I doing these correctly?
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  12. #11  
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    Yeah, these all look good to me.
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  13. #12  
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    All right! Now how about these four?

    #13 to #16 The given curve is rotated about the y axis. Find the area of the resulting surface.

    13) y = x<sup>1/3</sup>
    1 < y < 2

    x = y<sup>3</sup>
    x' = 3y<sup>2</sup>
    (x')<sup>2</sup> = 3<sup>2</sup>y<sup>4</sup>

    A = ∫<sub>1</sub><sup>2</sup> 2πy<sup>3</sup>(1 + 9y<sup>4</sup>)<sup>1/2</sup> dx

    14) y = 1 - x<sup>2</sup>
    0 < x < 1

    y' = -2x
    (y')<sup>2</sup> = 4x<sup>2</sup>

    A = ∫<sub>0</sub><sup>1</sup> 2π(1 - x<sup>2</sup>)(1 + 4x<sup>2</sup>)<sup>1/2</sup> dx

    15) x = (a<sup>2</sup> - y<sup>2</sup>)<sup>1/2</sup>
    0 < y < (a/2)

    x' = -y(a<sup>2</sup> - y<sup>2</sup>)<sup>-1/2</sup>
    (x')<sup>2</sup> = y<sup>2</sup> / (a<sup>2</sup> - y<sup>2</sup>)

    A = ∫<sub>0</sub><sup>(a/2)</sup> 2π(a<sup>2</sup> - y<sup>2</sup>)<sup>1/2</sup>( 1 + y<sup>2</sup> / (a<sup>2</sup> - y<sup>2</sup>) )<sup>1/2</sup> dx

    16) x = a cosh (y/a)
    -a < y < a

    x' = sinh (y/a)
    (x')<sup>2</sup> = sinh<sup>2</sup> (y/a)

    A = ∫<sub>-a</sub><sup>a</sup> 2π( a cosh (y/a) )(1 + sinh<sup>2</sup> (y/a) )<sup>1/2</sup> dx

    Are these correct?
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  14. #13  
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    Yeah, this all look good to me, too. In each case, not only am I seeing whether you applied the formula correctly, but I'm also double checking by thinking about how I'd go about integration. If it's possible, then I figure that must be the right answer, because as you observed, these integrals are often really hard or even impossible to solve.
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  15. #14  
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    Ok, so its back to these old physics problems again.

    1) An aquarium 5th long, 2 ft wide, and 3 ft deep is full of water. Find (a) the hydrostatic pressure on the bottom of the aquarium. (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the aquarium.

    My work (1a):

    F = pgAd
    p = density, kg/m<sup>3</sup>, water desity is 1000 kg/m<sup>3</sup>
    g = gravity, m/s<sup>2</sup>, normally gravity is 9.8 m/s<sup>2</sup>
    A = area of object that is pressure is applied to, m<sup>2</sup>
    d = depth of object under liquid surface, m

    Ok, lets just work on some basic ideas of the first one first. First, am I right in thinking that since we are using the American system of measurments that I would not factor in density and gravity? Or would I convert them into feet and pounds? I went back and read a little of chapter 6 and it seems to leave out gravity in its book dropping example, and when it measures water it seems to measure the weight instead of mass.
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  16. #15  
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    This is my attempt to answer my own question.

    We know that from the exercises we did in chapter 6, that the weight of water is 62.6 lbs/ft<sup>3</sup>. This is already the force that is applied to a cubic foot of water after the effects of gravity and mass are factored in. In short, when working with feet and pounds, pg = 62.5 lbs/ft<sup>3</sup>.

    With the SI system we can take both p and g individually. For water, p = 1000 kg/m<sup>3</sup>, and g = 9.8 m/s<sup>2</sup>. If you take these together you get the SI equivilant pg = 9800 N.

    So if we go back to our problem 1:
    1a) The forumla for hydrostatic pressure is
    P = pgd

    We know that pg = 62.5, and d is given as 3 ft, so the hydrostatic pressure is
    375/2 pounds

    1b) The forumla for hydrostatic force is
    F = pgAd

    pg = 62.5

    A = (5 ft)(2 ft)
    A = 10 ft<sup>2</sup>

    d = 3

    F = (62.5)(10)(3) pounds
    F = 1875 pounds

    1c) The formula for hydrostatic force is
    F = pgAd

    If we are working with one end of the tank, then our d should vary, and we would have different values of F at different depths, so the best way to find this would be to set up an integral that accounts for this. This is the integral that I set up:
    ∫<sub>0</sub><sup>3</sup> (pgAd) dx
    ∫<sub>0</sub><sup>3</sup> (62.5)(2x)(x) dx

    I've also tried:
    ∫<sub>0</sub><sup>3</sup> (62.5)(6)(x) dx

    both these are wrong, so I've attempted to do something more simple. I tried to find an integral that represents an area. I chose an object that has dimensions 2 by 3, so the area should be 6.
    ∫<sub>0</sub><sup>3</sup> 2x dx

    which turns out to be 9, not 6 as I had hoped.

    What are the errors that I have made in my work above?
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  17. #16  
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    After thinking about it a bit more, I'm thinking that the dx end of the integral represents the change of x from the lower limit to the upper limit. If we did something simple such as find the area of an object that is 2 by 3, we could set up an integrand that looks like this:
    ∫<sub>0</sub><sup>3</sup> 2 dx

    This would give the area of 2 multiplied by x from 0 to 3, which would be 6.

    If we go back to our force integrand of 1c, then using this information I would change this integrand to:
    ∫<sub>0</sub><sup>3</sup> pgA dx
    ∫<sub>0</sub><sup>3</sup> (125/2)(10) dx
    ∫<sub>0</sub><sup>3</sup> 625 dx
    625[x]<sub>0</sub><sup>3</sup>
    1875 ft-lbs

    which is still wrong.
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  18. #17  
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    Quote Originally Posted by Demen Tolden
    If we are working with one end of the tank, then our d should vary, and we would have different values of F at different depths, so the best way to find this would be to set up an integral that accounts for this. This is the integral that I set up:
    ∫<sub>0</sub><sup>3</sup> (pgAd) dx
    ∫<sub>0</sub><sup>3</sup> (62.5)(2x)(x) dx
    Hi Demen Tolden. (a) and (b) look right to me. For (c) you're right about the hydrostatic pressure varying with depth...so what you want is the integral of Pressure with respect to depth between x=0 and x=3. So you can ignore force for a moment and concentrate on pressure.

    If F=pgAx, then P = pgx, so taking the integral:

    <sub>0</sub>∫<sup>3</sup>pgx dx
    = [(1/2)pgx<sup>2</sup>]<sub>0</sub><sup>3</sup>
    =(1/2)*62.5*9
    =281.25 lbft<sup>-2</sup>

    To find the force, I would assume 2ftx3ft to be the dimensions of an 'end' of the pool so:

    F=PA
    =281.25*3*2
    =1687.5 lb

    I think thats right?
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  19. #18  
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    The book says that the answer we want to come up with for 1c is 562.5 lbs. Thanks for the help. I'll be working on this more tonight, and I'll post if I get any advancement on the idea.
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  20. #19  
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    hmmm...I was getting 562.5 cropping up in my calculation before but that was for the pressure, and before I remembered to multiply by the half.
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  21. #20  
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    Demen, I figured out my mistake...instead of having:

    F=281.25*3*2

    it should just be:

    F=281.25*2=562.5 lb

    ...since we don't need to multiply by the depth again, because the depth has already be taken into account (or rather the integral of pressure with respect to depth).
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  22. #21  
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    Yea, that makes sense. Thanks.
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  23. #22  
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    So, could we say that the formula for finding the hydrostatic force against a submerged two dimentional rectangle would be:
    F = ∫<sub>a</sub><sup>b</sup> pgw dx

    where a is the distance from the liquid surface to the top of the submerged object, b is the distance from the liquid surface to the bottom of the submerged object, and w is the width of the rectangle?
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  24. #23  
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    Quote Originally Posted by Demen Tolden
    So, could we say that the formula for finding the hydrostatic force against a submerged two dimentional rectangle would be:
    F = </sup> pgw dx

    where a is the distance from the liquid surface to the top of the submerged object, b is the distance from the liquid surface to the bottom of the submerged object, and w is the width of the rectangle?
    I know I'm interuppting here, but I was just wondering exactly what this:

    ∫<sub>a</sub><sup>b

    symbol means? I'm at a wall with some of my theory calculations and algebra just can't cut it. I think its time I learned calculus .
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  25. #24  
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    This is an integrand. It is a symbol that is used to show an integral. An indefinite integral looks something like this:
    ∫ 4x dx

    If we evaluate this integral we end up with:
    2x<sup>2</sup> + C

    When I am typing out these problems I usually leave the C out even though it should always accompany an indefinite integral. The C represents all possible constants such as 2, or -66, or π, or whatever, as long as there is no variable attached to it.

    A definite integral may look like this:
    ∫<sub>a</sub><sup>b</sup> 4x dx

    If we evaluate this integral we end up with:
    [2x<sup>2</sup>]<sub>a</sub><sup>b</sup>
    2b<sup>2</sup> - 2a<sup>2</sup>

    Integrals can be used for many useful things. Pretty much any problem you want to solve by taking an infinitly large number of infinitly small portions you can solve using these neat little things. Some examples are areas between curves, volumes of objects that are not so ordinary looking, physics problems of force, or work, or pressure. They can be used for pretty much anything. Integration though requires skills of differentiation, so if you want to learn calculus you should start with ideas of functions and inverse functions, then limits, then differentiation, and so on.
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  26. #25  
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    Thank you Dementolden!
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  27. #26  
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    As Demen says, you must look at functions and limits first, and then learn differentiation, then integration, and do alot of practise questions too. I would also say that for a good understanding of higher science (whether that be physics/biology/geological sciences...whatever), calculus is a necessity, not just a bonus.
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  28. #27  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Is this right?

    2) A swimming pool 5 m wide, 10 m long, and 3 m deep is filled with seawater of density 1030 kg / m<sup>3</sup> to a depth of 2.5 m. Find (a) the hydrostatic pressure at the bottom of the pool, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the pool.

    My work:
    2a) P = pgd
    P = (1030 kg<sup>1</sup> m<sup>-3</sup>)(9.8 m<sup>1</sup> s<sup>-2</sup>)(2.5 m<sup>1</sup>)
    P = 25235 kg<sup>1</sup> m<sup>-2</sup> s<sup>-2</sup>

    I guess I just feel uncomfortable with units at the moment.

    2b) F = pgAd
    F = (25235 kg<sup>1</sup> m<sup>-2</sup> s<sup>-2</sup>)(50 m<sup>2</sup>)
    F = 1261750 kg<sup>1</sup> s<sup>-2</sup>

    2c) F = ∫<sub>0</sub><sup>2.5</sup> pgwx dx
    F = ∫<sub>0</sub><sup>2.5</sup> (1030 kg<sup>1</sup> m<sup>-3</sup>)(9.8 m<sup>1</sup> s<sup>-2</sup>)( x meters as a dimension of the area)(dx meters of varying depth)
    F = ∫<sub>0</sub><sup>2.5</sup> 10094 x dx (in units of kg / s<sup>2</sup>)
    F = 5047 [x<sup>2</sup>]<sub>0</sub><sup>2.5</sup>
    F = (5047)(6.25)
    F = 31543.75 kg / s<sup>2</sup>
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  29. #28  
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    With regard to 2c, would not the hydrostatic pressure vary linearly with depth? If so, can you not simply take an average instead of having to integrate over the distance?

    Plus it looks as thought there must be something wrong with 2a, because I've never heard of pressure being measured per second squared. Perhaps it's just a units issue, or perhaps there's something missing (though I can't think what).
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  30. #29  
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    I believe that a pascal (Pa) is a unit of measure that measures newtons (N) per square meter. A newton is kg m / s<sup>2</sup>, so therefore a pascal should be kg m<sup>3</sup> / s<sup>2</sup>.

    Does that sound right?

    Maybe you are right, but I think that integrating would be like adding the pressure at all depths even though there are infinite depths. Since I am trying to understand integration, I am trying to get this idea down of how to construct an integral correctly to calculate these kinds of things.

    Thanks for the help.
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    Quote Originally Posted by sunshinewarrio
    With regard to 2c, would not the hydrostatic pressure vary linearly with depth? If so, can you not simply take an average instead of having to integrate over the distance?
    The hydrostatic pressure varies linearly with depth, but rather than an average, we are interested in the total pressure exerted on the side between the surface, and some depth, x, where the pressure is a function of depth. In other words the integral.

    Demen,

    a.) The formula, and answer are correct, but the units are slightly off. If we look at the dimensions and the forumla we can find out what the final unit will be.

    P = pgd = kgm<sup>-3</sup> * ms<sup>-2</sup> * m = kgm<sup>2</sup> / m<sup>3</sup>s<sup>2</sup> = kgm<sup>-1</sup>s<sup>-2</sup>

    it might help to derive the equations too:

    p = m / V
    m = pV

    V = Area*depth = Ad

    m = pAd

    Force (weight) = mg = pgAd

    Pressure = F/A = pgAd/A = pgd

    The units for force are kgms<sup>-2</sup> (easy enough), and you can say that kgms<sup>-2</sup> / m<sup>2</sup> = kgm<sup>-1</sup>s<sup>-2</sup>. We can of course just use Newtons for force, and Pascals for pressure (In SI units).

    b.) is right, but the units should be kgms<sup>-2</sup> (F=mg) or N.

    c.) looks right too. Did you use 'w' for width? Again the units are as above.
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    ...Why has that writing gone small??
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    I think your writting got small because of a typo after your s<sup>-2</sup>.

    Your < / sup > was mistyped.

    Thanks for the help. I see my mistake. I'm glad I got the integral right. I'm on my way to understanding integration as a visual thing.
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    Sorry for my absence! I've been a bit busy to visit the forum. It looks like you guys are getting good work done, though, and I'm not really familiar with the section you're working on right now.
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    I'm glad you are back! (If your business is subsiding.) If you are still busy hurry up!
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    Quote Originally Posted by Demen Tolden
    I think your writting got small because of a typo after your s<sup>-2</sup>.

    Your < / sup > was mistyped.

    Thanks for the help. I see my mistake. I'm glad I got the integral right. I'm on my way to understanding integration as a visual thing.
    Thanks, and glad to help....post editied...now if we had LaTeX these things wouldn't happen....

    Sorry for my absence! I've been a bit busy to visit the forum. It looks like you guys are getting good work done, though, and I'm not really familiar with the section you're working on right now.
    Yeh, glad your back too, hope you've got things sorted.
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  37. #36  
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    Think that this integral is set up correctly?

    5) Find the hydrostatic pressure on a half circle with a radius of 10 m where the origin is at the water surface.

    My work:
    F = ∫ (7<sup>2</sup>5<sup>1</sup>2<sup>2</sup> (pg) )(2<sup>1</sup>(5<sup>2</sup>2<sup>2</sup> - x<sup>2</sup>) (width) )( x (height) )( dx (depth) )

    I worked this out and got 7<sup>2</sup>5<sup>5</sup>2<sup>5</sup> N
    which is
    4.9 10<sup>6</sup> N

    The book answer is 6.5 10<sup>6</sup> N

    I apologize if the exponents make it confusing to read. I find that it is easier to work out integrals like these using exponents.
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    Is it looking for a pressure or a force?...a pressure can exert a force on a 2D surface only, and not on a 1 dimensional profile. If it wants the force, you'd have to know the length of the container too.... This is assuming the semi circle is a cross-section of a half cylinder....but it might be a hemisphere shaped bowl?...in either case you'd need an area to calculate the force.
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    Ok, I know the print might be a bit small, and this is kind of a mess to just throw in one post, but this is a graphical representation of the problems that I am working on. The book's description of the exercises:

    3 to 9) A vertical plate is submerged in water and has the indicated shape. Explain how to approximate the hydrostatic force against the end of the tank by a Riemann sum. Then express the force as an integral and evaluate it.

    I'm kind of skipping the Riemann sum part if that is okay.

    I also made a mistake in my last post that bit4bit pointed out. I meant the hydrostatic force, not pressure.
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    Ok. I'll just redo all seven of these. I've made a shortcut for pg for US Customary units where
    pg = 5<sup>3</sup>2<sup>-1</sup> lb / ft<sup>3</sup>

    and metric units where
    pg = 7<sup>2</sup>5<sup>1</sup>2<sup>2</sup> in units of kg<sup>1</sup> m<sup>-2</sup> s<sup>-2</sup>

    I had previously made an error on problem 5 where my pg was miscalculated.

    The formula for F is
    F = pgAd

    where A is the area of the object, d is the depth of the object, and with the US system, pg is the weight density while with the metric system p is the mass density and g is the gravity. Since we are integrating with respect to the depth, pg will most likely always remain the same no matter the depth, while dimensions of the area will change with respect to the depth. Thus much of the time we will have to show the area's width and height as something that looks like
    (x - 1) (x - 2) or (2x + 5) x

    where the first factor in parintheses is the width and the second is the height. A width facter such as (x-1) or (2x+5) would mean that the width changes as the depth changes. A height factor such as (x - 2) would mean that the object in question is submerged 2 units under the surface. A height factor of x would mean that the object is at the surface. The dx at the end of the integral represents the depth. Thus an integral for F may look as follows:
    F = ∫ pgA dx

    Now lets get to the exercises.

    3) F = ∫<sub>2</sub><sup>6</sup> 5<sup>3</sup>2<sup>-1</sup> 6(x-2) dx
    F = 5<sup>3</sup>3<sup>1</sup>2<sup>-1</sup>[(x-2)<sup>2</sup>]<sub>2</sub><sup>6</sup>
    F = 5<sup>3</sup>3<sup>1</sup>2<sup>3</sup> lb
    F = 3000 lb

    The book answer is 6000 lb. I've done this problem before and arrived at the correct answer, but looking back on my work, the integral I used to get the correct answer doesn't look right. Instead of using (x - 2) as the height as I did above, I used (x). The reason this doesn't make sense to me is that the object is 2 ft below the surface, so the height of the object at x=2 should be 0 and so on. The integral I set up above looks correct to me. Before I move on, what is the problem with it?
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    Hi Demen, I see what you mean about the submerged plates...Its asking you to find the hydrostatic force on the surface of the plates. I had some picture in my head of oddly shaped containers.


    I think for these problems, its much better to consider the pressure as a function of depth first, rather than force, and then substituting an area (It actually tunrs out to be a length) in at the end to obtain the force...remembering that F=P*A.

    If,

    P=pgx

    ...then pressure varies proportionally with depth. To find the total pressure over x (within an interval [a,b]) we take the definite integral of pgx w.r.t x, between a and b. However, we have only taken one dimension into account (x), and to obtain the pressure acting over the surface (The Force) we must take into account the other dimension, its width. For the case of a rectangular surface (like the container walls in you previous problem), you can simply multiply it by the width of the wall.

    These questions are more complicated, because the width actually varies with depth. To get an exact answer to these questions, we need use to double integrals in multivariable calculus, but the Riemann sum allows you a way to approximate the answers, and is the only single variable calculus method of finding them.

    Basically, picture the (linear) graph of pressure vs depth. When you integrate the function, you are finding the area beneath the graph, above the x-axis, and between two limits [a,b]. The limits [a,b] refer to the depth of the uppermost point of the object, b, and the depth of the lowermost point a. (notice a and b are 'inverted' because depth is the opposite of height).

    Next you have to partition the interval into a group of n smaller partitions (how many you chose is up to you, but more partitions = more accuracy). You can either use left point, right point, mid point, or trapezoidal areas, which is up to you, but in each one, you need a value for the height of each partition (from the x-axis to the graph)....(trapezoidal is more accurate but more work too )..If your stuck with this I'll explain it better.

    so basically you can now find the area of each column. The sum of these areas is an approximated area of what we were originally trying to find, the integral...i.e.:

    ∫<sup>a</sup><sub>b</sub>pgxdx ≈ΣA<sub>i</sub>

    where A<sub>i</sub> are the areas of each column between i=a, and i=b. What this now allows us to do, is to find an approximate force for each interval by muliplying in the widths of the shape AT THE CORRESPONDING DEPTHS. For any particular above partition, you can use the width at the depth of the beginning of the partition, the end of the partition or even the average....again if you want me to explain better I'm happy to.

    So now our expression becomes:

    F ≈ΣA<sub>i</sub>w<sub>i</sub>

    where w<sub>i</sub> are the widths corresponding to the depths. This problem can be quite hard to think about all in your head, and a diagram really helps to understand whats what...I skethced a couple out thinking about this problem, and it helped to see whats happening. Hope that helps
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  42. #41  
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    Quote Originally Posted by bit4bit
    ...then pressure varies proportionally with depth. To find the total pressure over x (within an interval [a,b]) we take the definite integral of pgx w.r.t x, between a and b.
    What do you mean by w.r.t x?

    Quote Originally Posted by bit4bit
    These questions are more complicated, because the width actually varies with depth. To get an exact answer to these questions, we need use to double integrals in multivariable calculus, but the Riemann sum allows you a way to approximate the answers, and is the only single variable calculus method of finding them.
    Since the width varies with depth, could we not use one variable to describe the depth and find a relationship with the same variable to describe the width, thereby keeping a single variable integral and staying within the realm of knowledge contained within my book?
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    sorry...w.r.t is sometimes used as an abbreviation of 'With Respect To'...when people cant be arsed to copy and paste symobols, and write the proper expression
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  44. #43  
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    This is an example using problem 4

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    Since the width varies with depth, could we not use one variable to describe the depth and find a relationship with the same variable to describe the width, thereby keeping a single variable integral and staying within the realm of knowledge contained within my book?
    I don't think so, because although pressure changes with depth P=f(x), and width changes with depth w=f(x), thw width doesn't change with pressure, nor the pressure with width. Really what we need to do to get an accurate answer is to integrate the pressure w.r.t. depth, between the uppermost and lowermost edges of the surface, and then to integrate this integral between the leftmost and rightmost edges of the surface...which would mean a double integral. But you can effectively do the same thing with a riemann sum by the method above...though you won't get an exact answer, but only an approximation.

    basically what you do is to slice the surface horizontally, into some number of partitions, each with a change in depth of Δx (they don't necessarily have to be of equal amounts of change, but this is the simplest case). On the graph of pressure versus depth, these will be columns bounded by the graph of P=f(x), the x-axis, and the upper and lower bound of the partition. You can approximate the area of each partition, by squaring it off at the top (rather than exactly filling the space under the curve)...and theres a few ways you can do it...All which result in the same thing to varying degrees of accuracy...an approximation of the area of the partition. Then by adding them up, you get an approximate area for the entire interval. Check out this link: http://www.math.hmc.edu/calculus/tut.../riemann_sums/

    The key is that with this numerical method, you have the opportunity to jump into the integration halfway through, and expand on it a little. I don't think there is any other single variable calc method of doing this to give an exact answer, but I could be wrong.
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    What if we take an infinite number of Riemann sums? That should be exactly the same as taking an integral.

    Pressure varies with depth, and

    P = pgd

    For our purposes in these problems pg is constant, so the only variable we have is d. The problem we want to address is to find the pressure against the area of the verticle plate, or

    F = PA

    the hydrostatic force. The problem arises that the pressure changes with depth, however if we make the height infinitely small by taking an infinitely large amount of area strips, this is no longer a problem. The solution becomes a matter of taking a limit. We could represent this in an integral like so:

    F = ∫<sub>a</sub><sup>b</sup> (pgx)(w)dx

    where a represents the top of the submerged plate, b represents the bottom of the submerged plate, (pgx) you could think of as the pressure where x is the depth, (w) is the width in terms of x, and dx represents the infinitly small height of the area strip.

    I think I've finally figured it out.

    Riemann sums where a subject in 5-1.
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    The answers to 3, 5, 7, and 9 are:
    3) 6000 lbs
    5) 6.5 10<sup>6</sup> N
    7) 3.5 10<sup>4</sup> N
    9) 1000gπr<sup>3</sup> N

    3) ∫<sub>2</sub><sup>6</sup> (125x/2 (pgx))(6 (w))(dx (height))
    ∫<sub>2</sub><sup>6</sup> (125x/2)(6) dx
    ∫<sub>2</sub><sup>6</sup> 375x dx
    375/2[x<sup>2</sup>]<sub>2</sub><sup>6</sup>
    375/2(36-4)
    6000 lbs

    I've got to stop here and get to work.
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    Nice one, looks like no. 3 is right. I still don't think you will be able to use that method for all the questions though due to the non-rectangular surfaces, and I'll try to explain why:

    What if we take an infinite number of Riemann sums? That should be exactly the same as taking an integral.
    Right!....in fact that is exactly the definition of a definite integral. Suppose we want to find the area bounded by the graph of some function f(x), the x-axis, and the lines x=a, and x=b. Using the riemann sum, we first partition the interval [a,b] into a number , n, of smaller partitions. Since a, and b are fixed, there is a finite difference between them, and so as n increases, the interval of each partition, Δx, decreases. More acuurately (for equal partitions):

    Δx = (b-a) / n
    n = (b-a) / Δx

    You will often have seen the limit for a definite integral expressed as n-->∞, rather than Δx-->0. Well from the relationship above, we can see why. As n tends to infinity, Δx must tend to 0 for it to be so. Therefore we can write the integral as follows:

    A≈∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)(x<sub>n+1</sub>-x<sub>n</sub>)

    For equal partitions:

    A≈∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx

    taking the limit:

    A=lim<sub>n-->∞</sub>∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx = ∫<sub>a</sub><sup>b</sup>f(x) dx

    Applying that to your problems, I'll use no. 5 as an example (semicircle):

    P = f(x) = pgx

    We know that the depth of the top of the object, a, is 0 at any point along the width, but the value of b is not fixed, rather it changes at different points along the width (this is the key to the problem)....or in other words we have a new function x=g(w) that bounds the lower edge of the semicircle.

    x<sup>2</sup>+w<sup>2</sup> = 100 (radius 10m)

    Re-arranging for x:

    x = √(100-w<sup>2</sup>)

    This is the new upper limit of integration, and so the new integral becomes:

    ∫<sub>0</sub><sup>√(100-w<sup>2</sup>)</sup> pgx dx

    Calculating the integral, you will see that you still have an unknown variable in the answer, w. Therefore, the only possible way we can now get the answer for this problem is to integrate again with respect to w, in the interval w=[-10,10]. Our expression now becomes:

    F = ∫<sub>-10</sub><sup>10</sup>[ ∫<sub>0</sub><sup>√(100-w<sup>2</sup>)</sup> pgx dx ] dw

    This is called a 'surface integral', and is usually written without the square brackets, as:

    ∫<sub>-10</sub><sup>10</sup>∫<sub>0</sub><sup>√(100-w<sup>2</sup>)</sup> pgx dx dw

    or even:

    ∫∫<sub>R</sub> pgx dx dw

    where R is the "domain of integration" (the semicircle).

    We are now getting into the realms of multivariable calculus, which is beyond the scope of your bok, but is nonetheless necesary to solve the problems exactly. As I said before though, the Riemann sum allows you to approximate the answers, within the realm of what you know. I'll try to explain what we have actually just done, to see why the Riemann sum makes it possible.

    Picture a three dimensional axes, with the z-axis for pressure, the x-axis for depth, and the y-axis for width. The function P= pgx, appears as a slanted plane crossing the x-yplane at the y-axis (representing the linear 'gradient' of the pressure with depth). Lying in the x-y plane, we have the domain of integration, R (the semicircle).

    Now, similar to how an integral of one variable finds an area below a function, the surface integral finds a volume below the surface, (In this case our surface is the plane P = pgx, but it could be any continuous function f(x,y) ) and above a domain in the x-y plane (in this case the semicircle, R). This volume is numerically equal to the force on the plate, and its not hard to see why. Effectively we have:

    Pressure*depth*width = Pressure*Area = Force

    ...a bit more extravagent of course due to the quirkiness of having non-rectangular domains, and a pressure gradient within the water.

    Similar to the definition of the indefinite integral earlier, we have to partition things and start adding them up prior to taking limits. I modified this pic, to make things clearer:



    First of all, we need the integral of pressure w.r.t depth. Picture the semicircle lying flat in the x-y plane.... this integral will be pictured as a vertical plane underneath the surface, above the domain, and parallel to the x-axis. We can approximate it's area using a Riemann sum. First, we have partition the domain with respect to x (indicating the horizontal slicing above) and then multiply each one by the distance at that point to the surface (from the x-y plane). Again we have:

    A≈∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx

    taking the lmit:

    A=lim<sub>n-->∞</sub>∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx = ∫<sub>a</sub><sup>b</sup>f(x) dx

    Next we partition with respect to the width (this will look like vertical slices on the pic above). We can then obtain an approximate volume, by muliplying each Δw, by the corresponding area at that point...which we have just worked out. We therefore get:

    V≈∑<sub>-10</sub><sup>10</sup>[ lim<sub>n-->∞</sub>∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx ] Δw

    This will look like a bunch of (m) vertical slabs within the volume we're trying to get. Taking the limit, we get this volume (the force):

    F = lim<sub>m-->∞</sub> ∑<sub>-10</sub><sup>10</sup>[ lim<sub>n-->∞</sub>∑<sub>a</sub><sup>b</sup> f(x<sub>n</sub>)Δx ] Δw

    = ∫<sub>-10</sub><sup>10</sup>∫<sub>0</sub><sup>√(100-w<sup>2</sup>)</sup> pgx dx dw

    Working through that, you can see that the volume cane be approximated using these steps:

    1.) Take Riemann sum of pressure w.r.t. x (your figure will now look like the one above)
    2.) Multiply each partition by its width. (calculate this using w=√(100-x<sup>2</sup>) in this case.
    3.) Add them together.

    ...effectively you are finding the force on each rectangle, and adding them together.

    Hopefully you can kind of see why Riemann sums are useful here....they allow us to jump into the integration "half-way through", allowing us to do things like this. Hope that helped.
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    Wow, thanks for that post.

    Help me to understand what the James Steward is trying to describe then and how your post relates:


    8.3 Applications to Physics and Engineering

    Among the many applications of interal calculus to physics and engineering, we consider two here: force due to water pressure and centers of mass. As with our prevous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and then evaluate the resulting integral.

    Hydrostatic Pressure and Force

    Deep-sea divers realize that water pressure increases as they dive deeper. This is because the weight of the water above them increases.

    In general suppose that a tin horizontal plate with the area A square meters is submerged in a fluid density p kilograms per cubic meter at a depth d meters below the surface of the fluid as in Figure 1.

    Figure 1

    The fluid directly above the plate has volume V = Ad, so its mass is m = pV = pAd. The force exerted by the fluid on the plate is therefore

    F = mg = pgAd

    where g is the acceleration due to gravity. The pressure P on the plate is defined to be the force per unit area

    P = F/A = pgd

    The SI unit for measuring pressure is newtons per square meter, which is called a pascal (abbreviation: 1 N/m<sup>2</sup> = 1 Pa). Since this is a small unit, the kilopascal (kPa) is often used. For instance, because the density of water is p = 1000 kg/m<sup>3</sup>, the pressure at the bottom of a swimming pool 2 m deep is

    P = pgd = (1000 kg/m<sup>3</sup>)( 9.8 m/s<sup>2</sup>)(2 m)
    = 19,600 Pa = 19.6 kPa

    An important principal of fluid pressure is the experimentally verified fact that at any point in a liquid the pressure is the same in all directions. (A diver feels the same pressure on nose and both ears.) Thus, the pressure in any direction at a depth d in a fluid with mass density p is given by

    P = pgd = δd

    Note: When using U.S. Customary units, we write P = pgd = δd, where δ = pg is the weight density (as opposed to p, which is the mass density.) For instance the weight density of water is δ = 62.5 lb/ft<sup>3</sup>.

    This helps us determine the hydrostatic force against a vertical plate or wall or dam in a fluid. This is not a stratforward problem because the pressure is not constant but increases as the depth increases.

    EXAMPLE 1 A dam has the shape of the trapezoid shown in Figure 2.

    Figure 2

    The height is 20 m, and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.

    Figure 3a

    SOLUTION We choose a vertical x-axis with origin at the surface of the water as in Figure 3(a). The depth of the water is 16 m, so we divide the interval [0,16] into sub-intervals of equal length with endpoints x<sub>i</sub> and we shoose x<sub>i</sub>* ∈ [x<sub>i-1</sub>, x<sub>i</sub>]. The ith horizontal strip of the dam is approximated by a rectangle with height Δx and width w<sub>i</sub>, where, from similar triangles in Figure 3(b),

    Figure 3b

    a/(16-x<sub>i</sub>*) = 10/20
    or
    a = (16 - x<sub>i</sub>*)/2 = 8 - (x<sub>i</sub>*)/2

    and so
    w<sub>i</sub> = 2(15 + a) = 2(15 + 8 - (1/2)x<sub>i</sub>*) = 46 - x<sub>i</sub>*

    If A<sub>i</sub> is the area of the ith strip, then

    A<sub>i</sub> ≈ w<sub>i</sub> Δx = (46 - x<sub>i</sub>*)Δx

    If Δx is small, then the pressure P<sub>i</sub> on the ith strip is almost constant and we can use Equation 1 to write

    P<sub>i</sub> ≈ 1000gx<sub>i</sub>*

    The hydrostatic force F<sub>i</sub> acting on the ith strip is the product of the pressure and the area:

    F<sub>i</sub> = P<sub>i</sub> A<sub>i</sub> ≈ 1000gx<sub>i</sub>*(46-x<sub>i</sub>*)Δx

    Adding these forces and taking the limit as n->∞, we obtain the total hydrostatic force on the dam:

    F = lim<sub>n->∞</sub> Σ<sub>i=1</sub><sup>n</sup> 1000gx<sub>i</sub>*(46 -x<sub>i</sub>*)Δx
    = ∫<sub>0</sub><sup>16</sup> 1000gx(46 - x)dx
    = 1000(9.8 ) ∫<sub>0</sub><sup>16</sup> (46x - x<sup>2</sup>) dx
    = 9800[23x<sup>2</sup> - x<sup>3</sup>/3]<sub>0</sub><sup>16</sup>
    =4.43 10<sup>7</sup> N

    EXAMPLE 2 Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft if the drum is submerged in water 10 ft deep.

    Figure 4

    SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that the origin is placed at the center of the drum. Then the circle has a simple equation. x<sup>2</sup> + y<sup>2</sup> = 9. As in Example 1 we divide the circular region into horizontal strips of equal width. From the equation of the circle, we see that the length of the ith strip is 2(9-(y<sub>i</sub>*)<sup>2</sup>)<sup>1/2</sup> and so its area is

    A<sub>i</sub> = 2(9 - (y<sub>i</sub>*)<sup>2</sup>)<sup>1/2</sup> Δy

    The pressure on this strips is approximately

    δd<sub>i</sub> = 62.5(7-y<sub>i</sub>*)

    and so the force on the strips is approximately

    δd<sub>i</sub> A<sub>i</sub> = 62.5(7 - y<sub>i</sub>*)2(9-(y<sub>i</sub>*)<sup>2</sup>)<sup>1/2</sup> Δy

    The total force is obtained by adding the forces on all the strips and taking the limit:

    F = lim<sub>n->∞</sub>Σ<sub>i=1</sub><sup>n</sup> 62.5(7-y<sub>i</sub>*)2(9-(y<sub>i</sub>*)<sup>2</sup>)<sup>1/2</sup> Δy
    = 125∫<sub>-3</sub><sup>3</sup> (7-y)(9-y<sup>2</sup>)<sup>1/2</sup> dy
    = (125)(7)∫<sub>-3</sub><sup>3</sup> (9-y<sup>2</sup>)<sup>1/2</sup> dy - 125∫<sub>-3</sub><sup>3</sup> y (9-y<sup>2</sup>)<sup>1/2</sup> dy

    The second integral is 0 because the integrand is an odd function(see Theorem 5.5.7). The first integral can be evaluated using the trigonometric substitution y = 3 sin Θ, but it's simpler to observe that it is the area of a semicircular disk with radius3. Thus

    F = 875∫<sub>-3</sub><sup>3</sup>(9-y<sup>2</sup>)<sup>1/2</sup> dy = (875)(1/2)π(3<sup>2</sup>)
    = (7875π)/2 ≈ 12,370 lb
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  50. #49  
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    I'm not familiar with the book your using, but I'll try and tackle this in the morning, when my mind is a bit fresher.
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    Hi, finally got access to this thread again.

    What Steward is describing is to firstly split the interval [top of surface, bottom of surface] into sub-intervals/partitions. The way that makes his method different to mine is that he considers the range of each sub-interval, Δx, to be small enough so that the pressure at the top of it (at x<sub>i-1</sub>) equals the pressure at the bottom of it (at x<sub>i</sub>). Then for a constant presssure you simply have F=PA...the area being given by the width of the partition times its height giving, F=PwΔx.

    He has then written the width as a function of the depth, x<sub>i</sub>* where: x<sub>i</sub>* ∈ [x<sub>i-1</sub>, x<sub>i</sub>]...meaning x<sub>i</sub>* is any value of depth within the sub-interval [x<sub>i-1</sub>, x<sub>i</sub>]. You then have:

    F=pgx<sub>i</sub>* f(x<sub>i</sub>*)Δx

    Where f(x<sub>i</sub>*) gives the width of the surface at the abitrary depth x<sub>i</sub>*. You can see how this is calculated in either case. 1.) from adding the extra width of the triangle at either end, using tanΘ=O/A...where the ratio O/A is constant in a right angled triangle with a set value of Θ. 2.) using the equation of a circle.

    He next adds up the forces on all the partitions, and takes the limit as n-->∞ giving the total force on the surface. I didn't think this was possible, but looking at this it certainly is, and looks like it gives the exact answer too. This looks like it might even work with more complicated functions too... for example if P=pge<sup>x</sup> rather than just P=pgx.

    Comparing this to my method, I was considering Δx to be relatively large, and was having to integrate the pressure over the partition, due to it still having a greater presure at the bottom than at the top....Stewards method is obviously more effective though...so don't listen to me :P. Mind you, surface integrals are effectively the same thing as this, and I would personally opt for using them over this in any application, because they feel more natural to use. You'll see what I mean if you carry on with calculus after this book...but until then this method is good. Sorry if my last post was a bit misleading.
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  52. #51  
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    Thanks for that! Man, I'm still real stuck on this idea. What is the difference between what I seem to understand and what Stewart is tyring to describe? I've thought that I have been applying the same idea, but there is obvoisly something I have been missing since I always seem to come up with an incorrect answer when the width varies.

    Oh, and I tried to get some extra help from another site and made a mammoth post that mostly fit in most of the information that I have already typed up, but I think also that the third section of that post might give some information of how I am doing these problems at this time. It is typed in latex, so it isn't exactly convertable to this site.

    http://www.mathhelpforum.com/math-he...-pressure.html
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    For number 4 try:

    Take Θ to be the angle at the lower point of the triangle.

    tanΘ = 4/3

    since Θ is constant:

    tanΘ= opposit / adjacent = 4/3 = w<sub>i</sub> / (4-x<sub>i</sub>)

    w<sub>i</sub> = 4(4-x<sub>i</sub>) / 3

    =16-4x<sub>i</sub> / 3
    =16/3 - (4/3)x<sub>i</sub>

    therefore:

    A<sub>i</sub>= (16/3 - (4/3)x<sub>i</sub>)Δx

    P<sub>i</sub>= pgx<sub>i</sub>

    F<sub>i</sub>= pgx(16/3 - (4/3)x<sub>i</sub>)Δx
    F<sub>i</sub>=(16/3)pgx<sub>i</sub> - (4/3)pgx<sub>i</sub><sup>2</sup>Δx

    F = lim<sub>n-->∞</sub>∑<sub>1</sub><sup>4</sup>(16/3)pgx<sub>i</sub> - (4/3)pgx<sub>i</sub><sup>2</sup>Δx
    F = ∫<sup>4</sup><sub>1</sub> (16/3)pgx - (4/3)pgx<sup>2</sup> dx
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  54. #53  
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    I basically did the same thing as you, but I don't trust my answers since I know that besides problem 3 I got the odd problems wrong.

    I expressed my width as (4 - (4/3)(x-1)) which is the same as your (16/3) - (4/3)x.
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    Well 4 - (4/3)(x-1) is definately equal to (16/3) - (4/3)x so that can't be wring. What did you do for the rest of it?

    Starting where I left off at the last post:

    F = ∫<sup>4</sup><sub>1</sub> (16/3)pgx - (4/3)pgx<sup>2</sup> dx
    F = [(16/6)pgx<sup>2</sup> - (4/9)pgx<sup>3</sup>]<sup>4</sup><sub>1</sub>
    F = (16/6)*62.5*16 - (4/9)*62.5*64
    F = 2666.67 - 1777.78 = 888.89 lb

    What was the answer supposed to be? Maybe there was something I did wrong? Or was this right and it was only the odd numbered questions you got wrong?
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    It is only the odd numbered questions that I know the answers for. I don't know what the answers are for the even problems, so the only way I have to judge their correctness is to compare the techniques I used with the odd numbers and see if it makes sense.

    I think you forgot to subtract your lower limit, meaning the 1. My work looks almost exactly the same

    5<sup>3</sup>3<sup>-1</sup>2<sup>1</sup> ∫<sub>1</sub><sup>4</sup> 4x - x<sup>2</sup> dx
    5<sup>3</sup>3<sup>-1</sup>2<sup>1</sup> [2x<sup>2</sup> - 3<sup>-1</sup>x<sup>3</sup>]<sub>1</sub><sup>4</sup>
    5<sup>3</sup>3<sup>-1</sup>2<sup>1</sup> ( (2<sup>5</sup> - 2<sup>6</sup> 3<sup>-1</sup>) - ( 2<sup>1</sup> - 3<sup>-1</sup>) )
    5<sup>3</sup>3<sup>-1</sup>2<sup>1</sup> ( 32/3 - 5/3)
    5<sup>3</sup>3<sup>-2</sup>2<sup>1</sup>3<sup>3</sup> = 5<sup>3</sup>3<sup>1</sup>2<sup>1</sup>
    750 lbs
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    Oh sorry, I wrote it down as 0 the second time for some reason. Anyway, I put the 1 back in and got the same as you.

    For number 5:

    w<sup>2</sup>+x<sup>2</sup> = 100
    w = (100-x<sup>2</sup>)<sup>1/2</sup>

    A = (100-x<sup>2</sup>)<sup>1/2</sup>Δx

    P=pg(10-x)
    =625-62.5x

    F=PA = (625-62.5x)(100-x<sup>2</sup>)<sup>1/2</sup>Δx

    F=lim<sub>n->∞</sub>∑<sub>0</sub><sup>10</sup>(625-62.5x)(100-x<sup>2</sup>)<sup>1/2</sup>Δx

    F=∫<sub>0</sub><sup>10</sup>(625-62.5x)(100-x<sup>2</sup>)<sup>1/2</sup>dx

    etc.. is that what you did?
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  58. #57  
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    What I did for the width on problem 5 was first figured the equation of the circle with the center of the circle at the origin was x<sub>1</sub><sup>2</sup> + y<sub>1</sub><sup>2</sup> = 10<sup>2</sup>

    It may seem a bit backwards first solving for y<sub>1</sub>, but I think it did actually work out well enough:
    y<sub>1</sub> = (+/-)(100 - x<sub>1</sub><sup>2</sup>)<sup>1/2</sup>

    We will only be interested in the negative y<sub>1</sub> though so we can make this equation into:
    y<sub>1</sub> = - (100 - x<sub>1</sub><sup>2</sup>)<sup>1/2</sup>

    We can make this relationship between y<sub>1</sub> and x (not x<sub>1</sub>), where x is the depth with this:
    y<sub>1</sub> = - x

    So our equation becomes:
    -x = - (100 - x<sub>1</sub><sup>2</sup>)<sup>1/2</sup>
    x = (100 - x<sub>1</sub><sup>2</sup>)<sup>1/2</sup>

    We can then reverse this around again to solve for x<sub>1</sub>:
    x<sub>1</sub> = (100 - x<sup>2</sup>)<sup>1/2</sup>

    and that equation gives us the coordinate of the circle edge if we know the depth. Remembering that our circle origin was in the middle of the circle, we know that the value of this coordinate is only half of the width, thus if we want the full width we have to multiply this equation by two. Doing so we get:
    2x<sub>1</sub> = 2(100 - x<sup>2</sup>)<sup>1/2</sup>

    or:
    w = 2(100 - x<sup>2</sup>)<sup>1/2</sup>

    I think our work deviated on this last step. My integral then looks like this:
    ∫<sub>0</sub><sup>10</sup> pgx 2(100 - x<sup>2</sup>)<sup>1/2</sup> dx

    solving this I get 490,000π, which is incorrect.

    Why do you say this:
    P=pg(10-x)
    Wouldn't that mean that the pressure is 0 at the bottom of the plate and increases as you approach the surface? Maybe you made a mistake in what you were trying to express, but what was that you were trying to do?
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  59. #58  
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    If you're using pg=62.5 lb/ft^2, then that's your problem. The given dimensions of the half-circle is 20m in diameter.. Notice it's METERS. You either have to change that to 65.xx ft (doing it in my head, don't know exact conversion) or change pg to 980 kg/m^2s^2... That should give you the right answer, I hope.[/code]
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  60. #59  
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    problem 4 was in US Customary units, problem 5 is in SI. I guess we may not have made this too clear. Using SI units on problem 5 I get 490000π.

    ∫ 9800x (which is the pgx) 2(100 - x<sup>2</sup>)<sup>1/2</sup> (which is the width) dx (which is the height)

    ∫ 19600(100 - x<sup>2</sup>)<sup>1/2</sup> dx

    x<sup>2</sup> = 10sin<sup>2</sup> Θ
    x = 10 sin Θ
    dx = 10 cos Θ
    sin Θ = x/10

    ∫ 19600(10)(cos Θ)10 cos Θ dΘ
    ∫ 1960000 cos<sup>2</sup>Θ dΘ
    ∫ 980000(1 + cos 2Θ) dΘ
    490000 ∫ 2 + 2 cos 2Θ
    490000 [2Θ + sin 2Θ ]<sub>0</sub><sup>10</sup>
    490000 [ 2 arcsin (x/10) + 2(x/10)( (100 - x<sup>2</sup>)<sup>1/2</sup> / 10 ) ]<sub>0</sub><sup>10</sup>
    490000( (π + 0) - (0+0) )
    490,000π N
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  61. #60  
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    Sorry Demen there was some mistakes in my last post as has already been pointed out...got the units mixed up, and wrote a couple of other things wrong. I'm not sure exactly what you've done there by trying to parameterize with Θ, but I re-did the problem and got something different. This is what I did:

    Consider the positive y axis acting downwards (or the semicircle and pressure gradient acting upwards). Picking an arbitrary depth, y<sub>i</sub> we know:

    P<sub>i</sub> = pgy<sub>i</sub>

    the area of the ith strip is given by:

    A<sub>i</sub> = w<sub>i</sub>Δy

    y<sub>i</sub><sup>2</sup>+x<sub>i</sub><sup>2</sup>=100
    x<sub>i</sub>=(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>
    w<sub>i</sub>=2(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>

    A<sub>i</sub>=2(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>Δy

    F<sub>i</sub> = P<sub>i</sub>A<sub>i</sub>
    =2pgy<sub>i</sub>(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>Δy
    =1000*9.8*2y<sub>i</sub>(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>Δy

    (using SI units)

    =19600y<sub>i</sub>(100-y<sub>i</sub><sup>2</sup>)<sup>1/2</sup>Δy

    Taking the limit of the sum:

    F=∫<sub>0</sub><sup>10</sup>19600y(100-y<sup>2</sup>)<sup>1/2</sup>dy

    Taking the substitution, u=100-y<sup>2</sup>:

    du/dy = -2y
    dy= (1/-2y)du

    but y= (100-u)<sup>1/2</sup> so:

    dy= (1/-2(100-u)<sup>1/2</sup>)du

    F= ∫<sub>100-0<sup>2</sup></sub><sup>100-10<sup>2</sup></sup>19600(100-u)<sup>1/2</sup>*u<sup>1/2</sup>*(1/-2(100-u)<sup>1/2</sup>)du

    F= ∫<sub>100</sub><sup>0</sup> [ (19600(100-u)<sup>1/2</sup>*u<sup>1/2</sup>) / (-2(100-u)<sup>1/2</sup>)du ]

    F= ∫<sub>100</sub><sup>0</sup>-9800u<sup>1/2</sup>du
    F= [-9800*(2/3)*u<sup>3/2</sup>]<sub>100</sub><sup>0</sup>
    F= [-6533u<sup>3/2</sup>]<sub>100</sub><sup>0</sup>
    F= 0 - [-6533*100<sup>3/2</sup>]
    F= 6533*1000 = 6.5MN

    It makes sense that these answers are much bigger, since the scale of the semicircle is bigger than in other questions, and we are using the correct SI units for it...since 1N is a small unit....However, I'm not sure whether my answer is too big.
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  62. #61  
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    Looking at your work, and comparing with mine, I realize that while evaluating my integral I somehow dropped the x out of my pgx. Putting it back in I come up with the same answer as you. Then looking back into the book I see that the answer for problem 5 is:
    6.5 10<sup>6</sup>

    I am in disbelief. I've been stuck on this for three weeks. Are the errors I have not in the forumluation of the integral like I was certain they were, but in simple little mistakes in the evaluation of the integrals?

    This is my work on problem 6:

    P = 1000gx
    w = 4x/5
    F = ∫<sub>0</sub><sup>5</sup> (1000gx)(4x/5)dx
    F = ∫<sub>0</sub><sup>5</sup> 800gx<sup>2</sup> dx
    F = (800g/3) [x<sup>3</sup>]<sub>0</sub><sup>5</sup>
    F = (800g/3)(125)
    F = 326667 N
    F = 326,667 N
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  63. #62  
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    Number six is right. 8)
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    Ok, well, for number 7, I'm sure this is set up right the way I understand it should, but I don't get the correct answer.

    7) P = pgx
    w = 12 + x
    F = ∫<sub>0</sub><sup>8</sup> (pgx)(12 + x) dx
    F = ∫<sub>0</sub><sup>8</sup> pg(12x + x<sup>2</sup>) dx
    F = pg[6x<sup>2</sup> + x<sup>3</sup>/3]<sub>0</sub><sup>8</sup>
    F = pg(384 + 512/3)
    F = pg(640/3)
    F = 13333 lb
    F = 13,333 lb

    Correct answer is:
    3.5 10<sup>4</sup> lb

    Nevermind, I see a mistake. I added (384 + 512/3) wrong. Fixing this I get 34,667 lb which seems to be correct.
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