# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=x^2 and y^2=x rotated about the x-axis?

Jun 19, 2015

To use shells, we take our representative slices parallel to the line we are revolving around. So the thickness will be $\mathrm{dy}$, the radius will be $y$.
The height will be the greater $x$ (the one on the right) minus the lesser $x$ (the one on the left). We need both curves with $x$ as a function of $y$.
The limits of integration will be $y$ values. That won't matter in this case, because the curves intersect at $\left(0 , 0\right)$ and $\left(1 , 1\right)$.

The two curves are: $y = {x}^{2}$ and $x = {y}^{2}$

Solving the first equation for $x$, we get $x = \pm \sqrt{y}$ which does not give $x$ as a function of $y$ (functions don't say "or"). Don't panic. A quick sketch of the curves shows us that the bounded region has only positive $x$ values, so we will use the function: $x = \sqrt{y}$.
The second equation already gives $x$ as a function of $y$.
Looking again at our sketch of the region, we see that the curve $x = \sqrt{y}$ is on the right and $x = {y}^{2}$ is on the left.
The height is $\sqrt{y} - {y}^{2}$

The volume of the representative shell is:
$2 \pi r h \cdot \text{thickness} = 2 \pi y \left(\sqrt{y} - {y}^{2}\right) \mathrm{dy}$
And the limits of the region are $y = 0$ to $y = 1$.

$V = {\int}_{0}^{1} 2 \pi y \left(\sqrt{y} - {y}^{2}\right) \mathrm{dy} = 2 \pi {\int}_{0}^{1} \left({y}^{\frac{3}{2}} - {y}^{3}\right) \mathrm{dy}$

I am sure that you can finish from here.