Integrating areas to find volumes

[AmySaunders]

Find the volume V of the described solid S.The base of S is a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares.




I get (64r^3)/3 as my answer, but WebAssign says it is incorrect. What am I doing wrong!?!?!?!!!

 

[HallsofIvy]

Since you don't say what you did, it is impossible to say what you did wrong! Yes, \(\displaystyle \frac{64r^3}{3}\) definitely wrong. (The "\(\displaystyle r^3\)", at least, is right. Since r is a length, a volume has to have units of length cubed.)

 

[AmySaunders]

First, I drew what I thought I was looking for, which was difficult, and may be incorrect. I think it looks like a circular base, with a sort of rectangular convex top on it, in order to have square cross sections, parallel to a diameter.

Second, to find volume, I want to integrate the area of the squares, from 0 to 4r, then doubled.

Third, to find area of each square, I used y=sqrt(r^2-x^2) and found 2*sqrt(r^2-x^2) to be the length of each side. So the area of each square is (2(sqrt(r^2-x^2)))^2= 4(r^2-x^2)

Fourth, I integrated from 0 to 4r, (r^2-x^2)dx, multiplied by an 8 in front of the integral, which is the 4 factored out, doubled, since I'm only integrating half the shape.

Thus the answer 64r^3/3.

Can you see where I'm going wrong now?

 

[HallsofIvy]

First, I drew what I thought I was looking for, which was difficult, and may be incorrect. I think it looks like a circular base, with a sort of rectangular convex top on it, in order to have square cross sections, parallel to a diameter.

Second, to find volume, I want to integrate the area of the squares, from 0 to 4r, then doubled.

Third, to find area of each square, I used y=sqrt(r^2-x^2)

This is your problem. Since the circle has radius 4r, it has equation \(\displaystyle x^2+ y^2= 16r^2\) so \(\displaystyle y= \sqrt{16r^2- x^2}\).

and found 2*sqrt(r^2-x^2) to be the length of each side. So the area of each square is (2(sqrt(r^2-x^2)))^2= 4(r^2-x^2)

Fourth, I integrated from 0 to 4r, (r^2-x^2)dx, multiplied by an 8 in front of the integral, which is the 4 factored out, doubled, since I'm only integrating half the shape.

Thus the answer 64r^3/3.

Can you see where I'm going wrong now?

 

[AmySaunders]

That was it, thank you!