[SOLVED] Volume of Solid Of Revolution....y=sqrt(x) and y=x, around x=5

Hello,

I have been trying to solve the following problem:

Consider the solid obtained by rotating the region bounded by the given curves about the line *x* = 5. The curves are: $\displaystyle y=\sqrt{x}$ and $\displaystyle y=x$. Find the volume *V* of this solid.

Facts: The only points of interesction are at (0,0) and (1,1).

Attempted Solution:

We need to use the washer method. However, since the region is being rotated around the line $\displaystyle x=5$, the radius of each circle will be $\displaystyle r = r + 5$. We also need to integrate from 0 to 1 along y. Therefore, my solution looked something like this:

$\displaystyle V = pi \int (y^2+5)^2dy - pi \int (y+5)^2dy$

expanding....

$\displaystyle V = pi \int y^4+10y^2+25dy - pi \int y^2+10y+25dy$

simplify...

$\displaystyle V= pi \int y^4+9y^2-10y dy$

integrate...

$\displaystyle V = pi(\frac{y^5}{5} + 3y^3 - 5y^2)$

$\displaystyle V = pi(\frac{1}{5} + 1 - \frac{5}{2})$

$\displaystyle V = \frac{-21}{10}pi$

The problem with my solution is that volume can not be negative. Can some one please explain what is wrong with my solution?

Thanks,