Find the volume of a solid...

[hank]

Find the volume of the solid that results when the region enclosed by x = y^2 and x = y is revolved about the line y = -1.

Here's my setup:

V = pi S [0 - 1] [sqrt[x] + 1]^2 - x^2 dx
= pi S [0 - 1] (x + 2sqrt[x] - x^2)
=pi (1/2 * x^2 + 4/3 * x^(3/2) + x - 1/3 * x^3) [0 to 1]
= 5pi / 2

However, the book shows the answer should be pi/2.


I reasoned the top function should be y - (-1) to compensate for the axis of revolution dropping one unit:

R(x) = y + 1 = sqrt(x) + 1

And the bottom function:

r(x) = y = x.

Can someone see where I went wrong?

 

[galactus]

Try shells:

\(\displaystyle \L\\2{\pi}\int_{0}^{1}(y+1)(y-y^{2})dy\)


Washers:

\(\displaystyle \L\\{\pi}\int_{0}^{1}\left[(-1-\sqrt{x})^{2}-(-1-x)^{2}\right]dx\)

 

[hank]

We're covering shells tomorrow.

However, with washers you've pushed me in the right direction.

However, I would have thought it should be (sqrt[x] + 1)^2 and (x +1)^2.

Why are your signs opposite?

 

[galactus]

Same thing.

 

[hank]

Same thing.

Ya, I just worked the problem and duh found that out.

It's squared, so it doesn't matter.

I got the problem. I didn't adjust for the movement of the graph for the bottom function.

Thanks!