Use the method of cylindrical shells to find the volume generated by rotating the region bounded...

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves $y=x^2, y=2-x^2$ about the line $x=1$.

I'm just trying to set this one up.

Is this correct?

$$2\pi\int_0^1 ((2-x^2) - (x^2))(1-x) dx$$

• That looks good! Commented Mar 31, 2015 at 3:31

Close. It is a very good thing to exploit symmetry, but here there is a problem. The region $A$ between the curves, from $x=-1$ to $x=1$ is congruent to the region $B$ between the two curves, from $x=0$ to $x=1$. However, when we rotate about $x=1$, Region $A$ sweeps out a much larger volume than Region $B$. So we cannot just rotate Region $B$ and double the result.

The fix is easy. By the Shells procedure that you know well, the volume is $$\int_{-1}^1 \pi\left((2-x^2)-x^2\right)(1-x)\,dx.$$

Remark: The integration is quite simple. We don't really need to worry about the terms obtained from the $-x$ part of $1-x$, since they integrate to $0$.

You specified Shells, but we also get a simple integral if we take slices perpendicular to the $y$-axis and integrate the cross-sectional area with respect to $y$.