# Find the volume of the solid of revolution

I have 2 lines, $y=x^2$ and $x=y^2$, and I am trying to solve for the volume of the solid created by rotating the region bounded by those 2 lines around the line $x=-1$.

The region bounded by these 2 lines looks somewhat like this (closest image I could find):

This region is then rotated around the line $x=-1$ which creates a bowl-like solid.

I know how to solve for the volume using $\int^b_a f(x)-g(x)\,\mathrm{d}x$, but I am getting tripped up by the fact that these two lines are functions of x and y. This is what I tried:

\begin{align*} V &= \int^1_0 \pi(\sqrt y)^2-\pi(y^2)^2\,\mathrm{d}y\\ V &= \pi\int_0^1 y - y^4\,\mathrm{d}y\\ V &= \pi[\frac{1}{2}y^2-\frac{1}{5}y^5]|_0^1\\ V &= \frac{1}{2}\pi - \frac{1}{5}\pi\\ V &= \frac{3}{10}\pi \end{align*}

Which is incorrect. What am I doing wrong?

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The inner and outer radii aren’t $\sqrt{y}$ and $y^2$, because your axis of revolution isn’t the $y$-axis. The axis of revolution is at $x=-1$, one unit further away than your calculation makes it. HINT: If you were using shells instead of washers, the radius of the shell at $x$ would be $x-(-1)=x+1$, not $x$. The method of washers also requires an adjustment, though the details are different. –  Brian M. Scott Feb 9 '12 at 5:08
So I have to add one to each of those?: $\sqrt y + 1$ and $y^2 + 1$? –  Marlon Feb 9 '12 at 5:10
Yes, before you square them. –  Brian M. Scott Feb 9 '12 at 5:11
@BrianM.Scott Wow thank you! I got the correct answer! If you make your comment an answer I will select it. –  Marlon Feb 9 '12 at 5:18

The inner and outer radii aren’t $\sqrt{y}$ and $y^2$, because your axis of revolution isn’t the $y$-axis. The axis of revolution is at $x=-1$, one unit further away than your calculation makes it.
HINT: If you were using shells instead of washers, the radius of the shell at $x$ would be $x-(-1)=x+1$, not $x$. The method of washers also requires an adjustment, though the details are different.