4. "The base of a solid object is a circle with a radius of 4. Every verticle cross section is an equilateral triangle. What is the volume of the solid?"

Ok I'm not sure about this 1, can some1 tell me if I'm doing it right

area of triangle = (1/2) x base x height

so I take the integral of the area of the triangle

the formula for a circle with radius 4 is (x'2/16) + (y'2/16) = 1

solved for y which is y = sqrt(16-x'2)= base

To solve for the height remembered that the cross section is an equilateral triangle

so all sides are equal to 8 since the radius is 4,

then divided the equilateral triangle in half to make 2 right angle triangles

so using a'2 + b'2 = c'2

4'2 + b'2 = 8'2

b = 4sqrt3

side b of the triangle is the height

so the intergal is {4 integral (2sqrt3)sqrt(16-x'2)dx} with an upper bound of 4 and a lower bound of 0

I put a 4 infront of the integral since the intergal is only solving for the 1st quadrent and I need to take account for the other quadrents.

Am I doing this right and can someone post the solution please. Thanks