Thread: Chapter 6: Applications of Integration

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Chapter 6: Applications of Integration

5) Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

y = x + 1
y = 9 - x²
x = -1
x = 2

My work:

I'll skip image posting and just describe the graph.
y = x + 1 has a slope of 1, a y-intercept of 1, and an x-intercept of -1.
y = 9 - x² is a downward turned parabola with a critical point at (0,9), a y-intercept of 9, and two x-intercepts of 3 and -3.
These two lines intersect at -3.37 and 2.37, and if we integrate with respect to x and use -1 < x < 2 (or equal to of course) to restrict the x values of our area, then these points of intersection are outside of our restricted area and of no consequence. We can then set up our integral as
∫_-1² [ (9-x²) - ( x + 1 ) ] dx
or
∫_-1² ( -x² - x + 10 ) dx
[-x³/3 - x²/2 + 10x]_-1²
(-8/3 - 2 + 20 ) - (1/3 - 1/2 -10 )
-3 - 3/2 + 30
51/2 or 25.5

The book answer is 19.5

I see my small silly mistake. I added wrong within the integral. Silly me.

A minor mistake at most--the rest of your work is perfect!

Ok, how about this problem from 6.2:

8 ) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y = sec x, y = 1, x = -1, x = 1, about the x-axis

My work:
Since we are rotating about the x-axis we can constrain the values of x to -1 < x < 1 (or equal to of course). We can set up our area function as
A = πr².

Since the line y = sec x is greater than or equal to y = 1 for all values of x between -1 and 1, our radius, r, would be equal to sec x - 1, or
r = sec x -1.

Substituting this into our area function we have
A = π ( sec x -1)².
or
A = π sec² x - 2π sec x + π

We can then set up our Volume integral as
V = ∫ (π sec² x - 2π sec x + π ) dx.

Integrating this I get
V = [π tan x + π x - 2π ∫ sec x dx]_-1¹,

but I can't quite figure out how to integrate ∫ sec x dx.

You don't quite have the right area function. Note that our solid is going to have a hole going down the middle, so the typical cross section is going to be a washer, not a circle. What's the formula for getting the area of a washer?

Just for fun, mathematicians usually call the washer an annulus.

Ok, so the area function of a washer is
A = π(outer radius)² - π(inner radius)²
so
A = π sec² x - π

then:
V = ∫_-1¹ (π sec² x - π)dx

= [π tan x - πx ]_-1¹
= (π tan 1 - π) - (π tan -1 + π)
= π tan 1 - π + π tan 1 - π
= 2π tan 1 - 2π
= 2π ( tan 1 - 1)

I don't seem to be able to get problem 9 right.

y² = x, x = 2y, about the y-axis.

A = π4y² - πy⁴
V = ∫₀⁴ ( π4y² - πy⁴ ) dx
= [4πy³/3 - πy⁵/5]₀⁴
= 256π/3 - 1024π/5
= -1792π/15

I know I shouldn't get a negative number, but I'm sure 2y gives a larger x value than y² when 0>y>4. The book answer is 64π/15. At least I got π/15 correct.

Shouldn't you be going from 0 to 2? And since you're integrating with respect to y, you should have dy as your differential.

You are right, my mistake.

What is wrong with my work on this 6.3 problem?

7) Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. Sketch the region and a typical shell.

y = 4(x-2)²
y = x² - 4 x + 7

My work:

y = 4(x-2)²
y = 4(x² - 4x + 4)
First Curve:
y = 4x² - 16x + 16

Second Curve:
y = x² - 4x + 7

Finding at what x values they intersect:
4x² - 16x + 16 = x² - 4x + 7
3x² - 12x + 9 = 0
x² -4x + 3 = 0
(x-1)(x-3) = 0
x = 1,3

Graphing shows y = x² - 4x + 7 to be greater than y = 4x² - 16x + 16 within [1,3]

Setting up a Volume integral:
V = 2π ∫₁³ x ( x² - 4x + 7 - (4x² - 16x + 16) ) dx
V = 2π ∫₁³ ( -3x³ + 12x² - 9x ) dx
V = 2π [-3x⁴/4 + 4x³ - 9x²/2 ]₁³
V = 2π ( -5/4 - (27/4) )

Another negative number

Oh, nevermind. I made another small mistake that I often make at the end. I subtracted the value when x = 3 from the value when x = 1.

Let me commend you on your ability to recognize a wrong answer and persevere in finding your mistakes. A surprising number of students don't see red flags when they get negative areas and volumes! Also, it's great that you can recognize patterns in the mistakes you make. All of these qualities are necessary to teach yourself mathematics.

Thanks.

I'm working on 6.4 now, and I've got a couple questions.

4) When a particle is located a distance x meters from the origin, a force of cos (πx/3) newtons acts on it. How much work is done in moving the particle from x = 1 to x = 2? Interpret your answer by considering the work done from x = 1 to x = 1.5 and from x = 1.5 to x = 2.

My work:
I've worked this out two ways. The first time I worked it out I just left the integral from x = 1 to x = 2, because I thought that the value from x = 1 to x = 2 should be the same as when you would break it up in half. This solution was 0. The next time I broke it in half and I pretty much got 0 as well. Why this is, I believe is that the work integral,
W = (3/π) ∫₁² (π/3) cos (πx/3) dx

works out to be
(3/π) [sin (πx/3) ]₁²

and sin (πx/3) has a local max at x = 1.5. Since 1.5 is in the middle of the interval of x, W will be 0. It doesn't make sense to me to use cos (πx/3) as a force function if the force should be positive.

-----------------------------------------------------------------
This next 6.4 problem I've worked out three different times three different ways, and got a different answer every time, none of which is the correct one. It doesn't seem like it should be that difficult of a problem either.

9) Suppose that 2 J of work is needed to stretch a spring from its natural length of 30 cm to a lenth of 42 cm. How much work is needed to stretch it from 35 cm to 40 cm?

My work:
2 = kx
2 = 0.12k
k = 50/3
f(x) = 50x/3
W = ∫_0.05^0.1 50 x dx / 3
W = 25/3 [x²]_0.05^0.1
W = 25/3 ( 1/100 - 1/400 )
W = 25/400
W = 1/16 J

The book answer is 25/24 J

I think I know why my problem 9 is wrong. The reason why is that in the equation
f(x) = kx

f(x) is a force function and not a work function. 2 J is the amount of work, not the amount of force. So instead we can set up the work integral
2 = ∫₀^0.12 f(x) dx

so
2 = F(0.12) - F(0)

where F(x) is the antiderivitive of f(x). However, I still don't know how I could work this problem out with this information.

Ok, so now I've substituted kx for f(x) within the integral. Here is my work:

2 = ∫₀^(3/25) kx dx
2 = [ kx² / 2 ]₀^(3/25)
2 = 9k / 1250
k = 2500/9

W = ∫_(1/20)^(2/20) 2500x dx / 9
W = 1250/9 [x² ]_(1/20)^(2/20)
W = 1250/9 ( 1/100 - 1/400)
W = 1250/9 ( 3/400 )
W = 25/24 J

/cheer

Good job figuring out number 9! A lot of problems in this book will turn the usual process of calculating integrals on its head. Instead of solving for the integral, you'll be given the value of the integral and some value in the integral--perhaps in the limits, perhaps in the definition of the function being integrated--will be left undetermined. In this case, as you did above, you'll just have to calculate the integral as usual with the unspecified values, set this equal to the known value for the integral, and then solve for the unknown value.

As for number 4, your work is correct, so from a mathematics standpoint, you've got it down. From a physical standpoint... well, you first have to accept that this book will take a lot of liberties with physics. So while this sort of force function might not make physical sense, we can consider it abstractly. Second, we have to think about what the force function means. In our examples of point particles, we're really thinking about them moving in one dimension--along the real number line. Then we can describe the force acting on them as a function of x. A positive force indicates a force moving in the positive direction. A negative force indicates a force moving in the negative direction. The magnitude (absolute value) of the force is always nonnegative, and the sign just tells us which direction the force is going.

17) A leaky 10-kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs 0.8 kg / m. Initially the bucket contains 36kg of water, but the water leaks at a constant rate and finished draining just as the bucket reaches the 12 m level. How much work is done?

My work:
I've set up an integral that looks like this:
W = ∫₀¹² (10 + 4x/5 + 36 - 3x) dx

Is this integral not correct?

I guess I needed to account for gravity. The answer I got multipied by 9.8 is the book answer.

That's essentially right. The only problem is you should be integrating a force function, not a mass function. To fix this, you just need to factor in the acceleration due to gravity.

Is this correct?

18 ) A 10 ft chain weights 25 lb and hangs from a ceiling. Find the work done in lifted the lower end of the chain to the ceiling so that it's level with the upper end.

My work:
W = ∫₀¹⁰ (5/4) x dx
W = 125 / 2

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

Originally Posted by Demen Tolden

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

I don't think so. The assumption here is that a perfectly efficient engine is used and you are just calculating the theoretical least amount of work. The integral, if there is one, comes in summing the work needed to account for the rise in height of infinitesimal 'slivers' of water from 0 (top of aquarium) to 0.5 (halfway down).

I'm guessing there's a cheating way to do it using normal algebra... but I could be wrong.

cheer

shanks

You are right sunshine. The integral is:
W = ∫₀^(1/2) (2000 kg*9.8 gravity * x) dx
W = ∫₀^(1/2) (19600x) dx
W = 2450 J

I'm not sure what I was doing wrong.

Originally Posted by Demen Tolden

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

Yes, I think you are right. If you siphoned the water out, it wouldn't require any energy. Matter of fact, you could run a turbine and get energy out. I think you have to make an asssumption that the pump discharge is at the top of the tank.

Originally Posted by sunshinewarrio

Originally Posted by Demen Tolden

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

I don't think so. The assumption here is that a perfectly efficient engine is used and you are just calculating the theoretical least amount of work. The integral, if there is one, comes in summing the work needed to account for the rise in height of infinitesimal 'slivers' of water from 0 (top of aquarium) to 0.5 (halfway down).

I'm guessing there's a cheating way to do it using normal algebra... but I could be wrong.

cheer

shanks

The cheating way would be like this. You have a mass of water that is .5*1*2=1 cubic meter and you are raising its center of gravity from 3/4 of the way up the tank to the top, which is .25 meter. So the work done is mgh = 1 cubic meter * 1000kg/cubic neter*9.8m/sec/sec*.25m=2450J

Originally Posted by Harold14370

Originally Posted by Demen Tolden

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

Yes, I think you are right. If you siphoned the water out, it wouldn't require any energy. Matter of fact, you could run a turbine and get energy out. I think you have to make an asssumption that the pump discharge is at the top of the tank.

Actually - you raise an interesting point there Harold (and Demen if you're still reading):

Is the answer to Demen's problem the same as the amount of work we could get out of the aquarium with a tap at the base, and allowing the water to run out (through a perfectly efficient turbine) until it was half full?

Originally Posted by sunshinewarrio

Originally Posted by Harold14370

Originally Posted by Demen Tolden

19) An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use the fact that the density of water is 1000 kg / m³.)

My work:
Don't you need to know more infornation about the method used to get the water out?

Yes, I think you are right. If you siphoned the water out, it wouldn't require any energy. Matter of fact, you could run a turbine and get energy out. I think you have to make an asssumption that the pump discharge is at the top of the tank.

Actually - you raise an interesting point there Harold (and Demen if you're still reading):

Is the answer to Demen's problem the same as the amount of work we could get out of the aquarium with a tap at the base, and allowing the water to run out (through a perfectly efficient turbine) until it was half full?

No because instead of moving the center of the slug of water up .25m you are moving it down .75 so your answer would be tripled.

For 18, I got the same answer as you! I'm just curious how you arrived at your integral, though. There are a million and one ways of setting these work integrals up, and so the meat of the problem is typically justifying that your integral represents the right thing.

I guessed. If
W = ∫_a^b f(x)dx
I kind of pictured a thought problem with myself grabbing the end of the hanging chain and realizing that there is hardly any effort necisary to lift the bottom link. This lead me to believe that at a, the start of the process, when the legth of the lifted chain is 0, that the weight of the chain is 0, and that as the bottom of the chain is lifted higher that the weight increases. When the bottom of the chain reaches the ceiling, the bottom has traveled 10 ft. Also according to my thought problem the weight of the chain if you were holding onto the bottom of the chain when it was at the ceiling should be half the total chain weight. Using this information I constructed a function that incorperates these proportionalities.
W = ∫₀¹⁰ (5/2 lbs per ft)(1/2 lift weight to total weight ratio)(x feet)dx

I guess I should be setting up force functions, but I think I may not understand enough to be able to do that.

Actually I really feel now like I understand this water pumping concept in terms of integrals of force functions, but I always seem to get the wrong answer. Here are three more examples:

21-24 ) A tank is full of water. Find the work required to pump the water out of the outlet. In excerses 23 and 24 use the fact that water wights 62.5 lb / ft³.

21) A diagram shows a triangle-based right prism with a flat top and a sharp bottom. The dimentions of the triangle-base are a base of 3 m, and a height of 3 m. The length from one base to the other is 8 m. A pump extends 2 m above the top of the prism.

My work:
I believe the volume of this object should be
(3 m triangle-base base)(3 m triangle-base height)(1/2)(8 m prism length)
or
36 m³

the work integral should then be
W = ∫₂⁵ (9.8 gravity)(36 m³ volume)(1,000 kg/m³ mass per volume)(x pump meters)dx
W = 3704400 J

Book answer: 1.06 times 10⁶ J

I'd write out my other two, but I've got to get to work.

Oh, I forgot to figure in that there is less distance for the more abundant water on the top to travel than the less abundant water on the bottom. The water level will drop much slower when the process begins, and faster when the process ends.

I am having difficulty finding a way to relate the change in width of a shape to the change of water level, which I currently believe to be the last important ingrediant to my understanding of these water draining problems.

21)


The integral I've come up with in which I've tried to factor in the idea that the larger water volume at the top of the container has less distance to travel than the lesser water volume at the bottom is
W = ∫₂⁵ (5 - x this is the width of the top layer of water)(8 meters length of top layer)(1000 kg of mass per cubic meter of water)(9.8 gravity)(x meters at which the water level is below the top of the pump) dx

This ends up giving an answer of 3175200 J, which is not correct.

I think you just made a math error somewhere. Could you show your work on the integration?

You are right. I did make a math error when multiplying 8*1000*9.8. Maybe I hit the wrong key accidently when calculating.

W = ∫₂⁵ (5 - x)( 8 )(1000)(9.8 )x dx
W = ∫₂⁵ (5 - x)(78400)x dx
W = 78400 ∫₂⁵ (5x - x²) dx
W = (78400) [5x²/2 - x³/3]₂⁵
W = 78400 ( ( 125/2 - 125/3 ) - ( 10 - 8/3) )
W = 78400 ( ( 375/6 - 250/6 ) - ( 60/6 - 16/6 ) )
W = 78400 ( 81/6 )
W = 1058400

Wow! Its the right answer!

but that was the easy one. I've been working on problem 22 for a long time today. The reason I was able to get problem 21 right was because I could see that the width of the surface of the water in the tank had a clear linear relationship with the distance from the top of the pump to the surface. In problem 22, this relationship has something to do with the equation for the circle x² + y² = (3/2)², but I just can't seem to figure it out.

22)

Draw a chord of the circle to represent the surface of the liquid and let x equal the distance from the top of the tank to the surface. Draw a right triangle with the radius R of the tank as the hypotenuse, and one side of the triangle is R-x. The third side is half the chord length and can be found by the pythogorean theorem. This is for the top half of the tank. A similar expression can be derived for the bottom half. I'd do the integration in two parts for the two halves of the tank.

You only need one integral--the expression you get will be the same for the top and bottom.

Hmm, when I woke up this morning and had my breakfast I was thinking about this problem, and after working with it a bit I figured out how to express the width of the water surface in terms of its distance to the top of the pump by first solving this circle equation for x
x² + y² = (3/2)²
x = ( 9/4 - y² )^1/2

and then finding an equation which relates y to u, the variable I will use for pump distance.
u + y = 2.5
y = 2.5 - u

substituting:
x = ( 9/4 - (2.5 - u)² )^1/2
x = ( 9/4 - (25/4 - 5u + u²) )^1/2
x = (-u² + 5u - 4 )^1/2

doubling for width
2x = 2(-u² + 5u - 4 )^1/2

Now that that is taken care of I set up an integral to solve the problem.
W = ∫₁⁴ (2(-u² + 5u - 4 )^1/2 m width)(6 m length)(1000 kg per m³)(9.8 gravity)(u surface distance to top of pump)du
W = 117600 ∫₁⁴ (5u - u² - 4 )^1/2 u du

However I don't believe I have the ability to integrate this.
z = 5u - u² - 4
dz = (5-2u)du

I don't know where I would get a 5 - 2u from to convert it into terms of z

You might need to be a little tricky with this one. Can you calculate:

₁∫⁴(5u-u²-4)^1/2du

somehow without using antiderivatives? If so, you can add and subtract (an appropriate multiple) of this quantity to your integral to get the (5-2u)du term that you desire.

I might be confused, but I thought that integration and finding the antiderivitive were basically the same thing. I've separated the polynomial into -1(u-1)(u-4) or (-u+1)(u-4) or (u-1)(-u+4), but I can't seem to get it. I've used different substitutions, z = -u+1, a = z² - 3z, b = -z -3 and I can't seem to find the right substitution. I'm not sure I know another way to integrate other than either just knowing what the answer is from the more obvious integrals or using substitution.

I'm saying that you can evaluate a definite integral without actually doing an antiderivative--for example, you can use geometry.

Ok, thinking about this geometricly, the graph of (5u - u² - 4)^1/2 seems to be an elipse with x-intercepts at 1 and 4 and critical points at x = 2.5. If I find the area of this elipse, is that going to help me solve this integral? Wow, this is so confusing. :P

To think up the answer to this problem I was thinking about the relationships integration and differentiation have. I'm so used to thinking of integration as solving for an area of a particular pair of lines, so then I thought... what would the process be called if I already knew the integral but wanted to find the original equation. An anti-area!

That was funny at least to me.

Okay, so (5x-x²-4)^1/2 looks like an ellipse. Set y equal to this quantity and square things out to see what sort of ellipse it is.

y² = 5x - x² - 4
y² + x² - 5x + 4 = 0

Well, its a circle it looks like with a center at ( 2.5 , 0 ).

Indeed it is!

Excellent! Now that we know that we know... hmm...

I don't believe I know how to integrate using geometry. I can find an area or a volume using an integral. I can find find the Work. I can solve an integral using substitution, but I don't think I know any more than this.

Originally Posted by Demen Tolden

y² = 5x - x² - 4
y² + x² - 5x + 4 = 0

Well, its a circle it looks like with a center at ( 2.5 , 0 ).

What's the radius?

1.5 of course.
and the area is 9π/4. The problem is that I don't see the relevance.

Okay, let me try an example. Suppose I wanted to find:

₀∫¹(-x²+2x)^1/2(-2x)dx

The thing that's preventing me from doing this is that I'd love to do substitution but can't. I'd love to set:

u = -x²+2x

But then I get:

du = (-2x+2)dx

And I don't see this floating around in my integral. If I were instead integrating:

₀∫¹(-x²+2x)^1/2(-2x+2)dx

I'd be in business, because I've got all the ingredients I need. This is equal to:

₀∫¹u du = u²/2 ₀|¹ = 1/2

What's the difference between these two integrals? Literally, it's:

₀∫¹(-x²+2x)^1/2(-2x+2)dx - ₀∫¹(-x²+2x)^1/2(-2x)dx = 2₀∫¹(x²+2x)^1/2dx

So if I have some way of calculating the last integral, I'm done. So then the question is... what do I know about the graph of y = (-x²+2x)^1/2? Well, this is:

y = (-x²+2x)^1/2
y² = -x²+2x
y²+x²-2x = 0
y²+x²-2x+1 = 1 ("completing the square")
y²+(x-1)² = 1

Second year high school algebra tells us that this is a circle with radius 1 centered at the point (1,0). I'm integrating from 0 to 1, so I'm calculating the area underneath the top arc of the circle, above the x-axis, and to the left of the line x = 1. This is a quarter of a circle, or π/4. Since I have:

KNOWN INTEGRAL - DESIRED INTEGRAL = CIRCULAR SECTOR

Then I have my desired integral is 1/2 - 2π/4 = (1-π)/2.

Ok, so

∫₁⁴ (5u - u² - 4)^1/2(5 - 2u)du = ∫₁⁴(5u - u² - 4)^1/2 ( 5 )du - (5u - u² - 4)^1/2( 2u ) du
∫₁⁴ (5u - u² - 4)^1/2(5 - 2u)du = ∫₁⁴(5u - u² - 4)^1/2 5 du - ∫₁⁴(5u - u² - 4)^1/2( 2u ) du
∫₁⁴ (5u - u² - 4)^1/2(5 - 2u)du = 5 ∫₁⁴(5u - u² - 4)^1/2 du - 2 ∫₁⁴(5u - u² - 4)^1/2 u du

I'm going to set M equal to the integral that we want to solve for.
∫₁⁴ (5u - u² - 4)^1/2(5 - 2u)du = 5 ∫₁⁴(5u - u² - 4)^1/2 du - 2 M
M = (5/2) ∫₁⁴(5u - u² - 4)^1/2 du - (1/2) ∫₁⁴ (5u - u² - 4)^1/2(5 - 2u)du

Now I'll start solving the second integral
M = (5/2) ∫₁⁴(5u - u² - 4)^1/2 du - (1/2) [(2/3)(5u - u² - 4)^3/2 ]₁⁴

Amazingly enough the second integral is 0
M = (5/2) ∫₁⁴(5u - u² - 4)^1/2 du - 0
M = (5/2) ∫₁⁴(5u - u² - 4)^1/2 du

So now is the part where I find the area of (5/2)(5u - u² - 4)^1/2 between the x values of 1 and 4 correct? Since (5u - u² - 4)^1/2 is a circle with radius 3/2 and a center on the x-axis, then the area above the x-axis would be
9π/8

and
M = (5/2)(9π/8 )
M = 45π/16

Is this correct?

Indeed it is! Good job! In chapter 7, you'll learn a method for doing this sort of integral without resorting to geometry called trigonometric substitution. It's not really prettier than the method you used, but it doesn't require the trick you had to use, and it can handle a wider range of integrals. In the end, to check your work, I plugged the function into a graphing utility and had my computer do the integration. If you have a graphing calculator or a utility on your computer, I'd suggest learning how to do calculus with them! In fact, I'd suggest getting one or the other in any case.

Then the answer for problem 22 is
W = 117600*45π/16
W = 330750π J

It surprises me that an integral that does not contain π in it has a solution that does.

I've aquired Mathematica. There were some other programs bundled up with it. I don't really know anything about any of them except what I've learned about Mathematica, and I don't think that is really much compared to everything the program can do. Most of what is in the help in the program seems to be beyond my mathematical understanding.

Ok, moving on to problem
24)


It doesn't seem like I can use the same concept here because I get to a point where the answer is discovered by solving two integrals that need to be taken apart, but the pieces kind of equal the same types of integrals. Soving for them seems to just put you in an infinite loop. Here is what I mean:

The integral that we need to solve for to find Work is
W = 125π/2 ∫₀⁵ (25x - x³)dx

I set this integral equal to M to make things simpiler.
M = ∫₀⁵ (25x - x³)dx

I find a related integral that I can solve for:
∫₀⁵ (25x - x³)(25 - 3x²)dx

I take that related integral integral apart:
∫₀⁵ (25x - x³)(25 - 3x²)dx = ∫₀⁵ (25x - x³)(25)dx - ∫₀⁵ (25x - x³)(3x²)dx

I substitute and simplify:
∫₀⁵ (25x - x³)(25 - 3x²)dx = 25 M - 3 ∫₀⁵ (25x - x³)(x²)dx

I evaluate the left integral:
0 = 25 M - 3 ∫₀⁵ (25x - x³)(x²)dx

I move M to one side of the equal sign:
25 M = 3 ∫₀⁵ (25x - x³)(x²)dx

I now have another integral that it seems I need to find in the same way as M. I will name this integral N and make a simple equation that might be helpful later when I find the value of N:
25 M = 3 N

So now I have to take apart N:
∫₀⁵ (25x - x³)(x²)dx

I find a related integral that I am able to integrate:
∫₀⁵ (25x - x³)(25 - 3x²)dx

I take that integral apart... Wait a second here, I just did this integral!

Regarding the problem you just posted, you don't need to resort to anything fancy to evaluate that integral--the integrand is just a polynomial, and we know how to do that straight up. Don't bring out the big guns unless the usual methods don't work. (This is one reason why you have to do so many exercises in calculus--you need to have intuition for whether a method will work or not.)

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It's good to be surprised that π appears in an integral that seems to only use "algebraic" things. By this I mean all the numbers and functions that appear in the integral can be obtained by algebraic operations (arithmetic operations and root extraction) on integers and the variable of integration. We'll never obtain π by twiddling around with rational numbers and algebraic operations. But if we throw calculus into the mix, we suddenly start getting interesting "transcendental" numbers (i.e., numbers which don't satisfy any polynomial with integer coefficients) popping up--first and foremost, e and π. All of a sudden, we start finding that geometric (or "analytic") quantities relating to algebraic objects (e.g., the unit circle is "algebraic" because it's given by x² + y² = 1) give us transcendental numbers (e.g., the area of the unit circle is π).

What this actually tells us is that the antiderivative of an algebraic function can be a "transcendental" function, i.e. something which cannot be expressed via algebraic operations. We've already seen this in a few cases:

ln'(x) = 1/x

arctan'(x) = 1/(1+x²)

These functions are still pretty special, though, because they satisfy "differential equations" involving algebraic functions. There are many theories (algebraic and "analytic") relating to such functions, none of which I know much about, but it's pretty fascinating stuff.

Of course! I get so used to using one technique that I get completely blind to something so simple sometimes.

W = 78125π/4 ft-lbs

I think those last 4 problems really helped a lot in my understanding of integrals. I did problem 25, and I got it wrong. I'm sure I must have just had a minor error in calcuation somewhere, because I feel I really do understand how to do it.

25) Supposed that for the tank in Exercise 21 the pump breaks down after 4.7 times 10⁵ J of work has been done. What is the depth of the water remaining in the tank?

My work:
I just set the the upper limit of the integral to a variable and solve for the variable. It should look something like this:

470000 = ∫₂ⁿ (2u-4 m width)(8 m length)(9.8 gravity)(1,000 kg per m³)(u m depth)du

I get n = 3.42547
so depth is 1.57453 m

book answer is
2 m

I'm sure I set up the integral right, I probably just make a minor miscalculation. I think I'm going to start working on 6.5 now.

I thought the width in 21 was 5-x. Where did the 2u-4 come from?

Damn, you are right.
u + x = 5
where u is the distance from the surface to the top of the pump, and x is the width.

so x, the width is 5 - u.

For some reason I thought that u + x = 2, and that x was only equal to half of the width. I guess I was still stuck on thinking about circles.

I may not understand the shape this question is describing.

Chapter Review 23) The base of a solid is a curcular disk with radius 3. Find the volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base.

Originally Posted by Demen Tolden

I may not understand the shape this question is describing.

Chapter Review 23) The base of a solid is a curcular disk with radius 3. Find the volume of the solid if parallel cross-sections perpendicular to the base are isosceles right triangles with hypotenuse lying along the base.

1. The only isosceles right triangle has other angles of π/4 (45 degrees). This is the first clue as to the determination of its shape. The maximum height of this object, given that the maximum length of hypotenuse will be 2r is = ?

2. If parallel sections taken perpendicular to the base are both isosceles right triangles then that's the next clue as to its shape: only one such section can have the maximum base width as its hypotenuse. So any section parallel to it must have a smaller hypotenuse (but is still an isosceles right triangle) thereby giving you a strong hint as to the rest of the shape of this object.

3. I will not pretend to have the mathematical skills to solve this one - but I'm hoping the hints about its shape will help...

cheer

shanks

I guess this is what is throwing me off.

"parallel cross-sections perpendicular to the base are isosceles right trianges with hypotenuse lying along the base."

So these cross sections are perpendicular to the base which implies a 90 degree angle. If the cross section and the base are both part of a triangle, then the hypotenuse must be adjacent to both sides, and opposite from the 90 degree angle. This mean that the hypotenuse can not be along the base.

Originally Posted by Demen Tolden

I guess this is what is throwing me off.

"parallel cross-sections perpendicular to the base are isosceles right trianges with hypotenuse lying along the base."

So these cross sections are perpendicular to the base which implies a 90 degree angle.

Possibly an error of presumption?

The perpendicular sections are perpendicular in the third dimension.

Hint: Think of the Sydney Harbour Bridge on top of a circle, with lines running straight and parallel (to each other) from the top of the bridge to the circumference of the circle. Looked at from above these lines will be perpendicular to the line of the top of the bridge.

I realize if the object is a disk, it will not have the open center that I drew.

check it

You're confused about what is meant by "base". You're correct in that base usually refers to the flat bottom of a geometric shape. However, you're thinking of the base of the triangle. You should be thinking about the base of the solid being described, and the description says the base of the solid is a circle. So think of a circle sitting on a table top, and then imagine the solid popping up from the base.

Originally Posted by serpicojr

check it

Zigackly. I was trying not to draw it for Demen, but perhaps your way was best...

I imagine that the integration problem would consist of summing up the triangles that vary by a trigonometric relationship. Too advanced for me, but I think that's the shape of the problem.

Agreed, it's often better to give hints instead of just giving in the outright answer. However, in this case, I felt that the vocabulary was rather confusing without a picture. And, given Demen Tolden's track record, I think he's the kind of guy who only needs to see something illustrated once before he can do it himself.

Yes, pictures help.
I always draw something out if I'm not sure what it looks like.

I think it's really important for this one particular problem that, for the moment, I don't get much help on it because to solve it I am constructing an integral without a previous template somewhere. Why am I posting my work then? I'm not sure. Maybe it will help later.

My work:
The first time around I figured the volume would be the integral of the base times the height, so I did
V =∫₀³ (π r² )(3/2) dr
V = 27π/2

which the book says is incorrect since it says the answer is 36. So I thought about why it was wrong, and of course! I calculated the volume of a cone, not this crazy object. So I needed a way to figure in this odd shape transition from circle to line. I instead thought I could calculate the volume of this object by cutting into four equal pieces strait down what I think of as the y-axis. Each of these four pieces would contain this right triangle, where at the y-axis the base of the 2-D triangle would be 3, and at a distance of 3 units away, the base would equal 0. I figured that this decrease in base and height could be represented by the formula
b = h = (9-x²)^1/2
which is derived from the forumla of the circle on the bottom of our 3D shape. Since I now could represent the base and height in terms of x, the variable representing the distance from the origin, I could create an area function in terms of x.
A = (1/2)(9-x²)
and now having an area function I could start constructing an integral for the volume. This is the integral I created:
V = 4 ∫₀³ (1/2)(9-x² area)(x units away from origin)dx
I worked this integral out to be 81/2, which is also wrong. I'll probably keep working on this at work.

Btw, thanks for the picture.

I like the quantity (9-x²): this represents the area of one of the triangular cross-sections. Since you're only calculating one quarter of the area, dividing by 2 and then integrating from 0 to 3 (as opposed to from -3 to 3) also makes sense to me. But why are you introducing the extra x? I know you bring it in to represent the number of units from the origin, but why should this factor into your integral? What does it represent geometrically?

I'd review the section on finding the volumes of solids, which I believe is 6.2. Emphasis is placed on the disc/washer method in this section, but there is a discussion about more general solids (such as the one we're working with now). There's a pretty general method (I may even call it a formula) for doing what we're doing right now.

I know this might sound kind of crazy, but I was visualizing a shrinking triangle moving away from the origin, and creating volume as it went. The two things that changed were the size of the triangle and its position along the x-axis. The definition in my head at the moment of an integral as it relates to volumes was to enter in the area...

Anyway, all that is wrong I see.
V = ∫_a^b A(x) dx

The sad thing is that I have done just about every problem in this chapter, and there is only 10 questions left until I've done nearly all of them. I think I need more practice doing different things with integration so I get used to being more versitile with it, and become a better problem solver in general with it. I need to have one question say "Find the average value of this function," and another to say" find the work done in draining this container," then the next to say "find the volume of this object if it is rotated about the y-axis." When the problems get clumped up and all the same, I forget what I learned two or three sections ago.

In the integral
V = ∫_a^b A(x) dx

does dx then represent the third dimension? That is what I believe at this time.

Your visualization of the problem (a shrinking triangle moving away from the origin) is spot on. And the area function of this triangle is just 9-x². And the answer to your question about dx is yes: if we consider the triangle to extend in the height and width dimensions, then the integral takes care of the depth dimension; we don't need to add anything to the integrand to take care of this.

I tell my students to think of integration as summation in these application sections. For the question at hand, I imagine a triangular slice as being a solid with height and width described by the triangle and depth equal to some infinitesimal quantity "dx". So (9-x²)dx represents the "volume" of this slice. Then the integral sign just represents adding up all of these slices. You can use similar intuition for work problems--you can calculate how far an infinitesimally thin slice of water has to move and then add over all the slices, or calculate how far an infinitesimally short piece of a chain has to move and then add over all the slices. Once you've set up the problem, the calculus takes over.

If your main confusion is setting up these application problems, I think you can probably jump ahead to chapter 7. This chapter is predominantly math-based, so you'll just be doing integrals in the wonderful vacuum of mathematical abstraction. If you still want to practice the applications, I'm sure it doesn't hurt to go back over problems you've already done. Making up your own problems (which is as easy as changing numbers, shapes of water tanks, ways of lifting chains, etc.) is easy, and you can pose them to us to check your answers.

I think what I am going to do is move on to the next chapter. If I want to come back and do these chapter 6 problems again, I can always do that.