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**VectorRun** So I'm suppose to find the volume of revolution of an area enclosed by $\displaystyle y=2-x^{2}$ and $\displaystyle y=x^2$ rotated about x=1.

So I'm using the cylindrical shell method $\displaystyle \int_a^b 2\pi rhdx$ and I'm integrating with respect to x; finding the integration points for the limits of integration, I get b=1 or a=-1.

Since the axis is x=1, then r=1-x. The height, $\displaystyle h=(2-x^{2})-(x^2)=2-2x^2$

Therefore, the definite integral is (factoring out the 2pi from the start and 2 from the height): $\displaystyle 4\pi \int_{-1}^{1} ((1-x)(1-x^2))dx$.

Since it's symmetrical I can change the limits to 0 and 1, multiplying the integral by a factor of 2. Then expanding yields: $\displaystyle 8\pi \int_{0}^{1} (1-x-x^2+x^3)dx$

By FTC: $\displaystyle =8\pi (x-\frac {x^2}{2}-\frac {x^3}{3}+\frac {x^4}{4})$ evaluated from 0 to 1.

That yields $\displaystyle 8\pi (1-\frac {1}{2}-\frac {1}{3}+\frac {1}{4})$ and simplifying gets: $\displaystyle \frac {10\pi}{3}$

However, I'm apparently wrong as the answer key says that it should be $\displaystyle \frac {16\pi}{3}$. I can't see my mistake, anyone see where I went wrong?