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?