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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?

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

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  • $\begingroup$ That looks good! $\endgroup$
    – AgentS
    Commented Mar 31, 2015 at 3:31

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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 $y$.

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