We see the function and a sketch of the volume in the following figure.
We use the washers method
![{\displaystyle V=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9d5758adf6b8e9f335a0b680278eb849893bf028)
to find the volume, where RO is the outer radius of the washer and RI is inner radius. From the diagram above, we see that
, and
.
We are integrating along x, so the boundaries of the integral must be
. Now we need to calculate the integral
![{\displaystyle {\begin{aligned}V&=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx\\&=\pi \int _{-1}^{1}((3-x^{2})^{2}-4)\,dx\\&=\pi \int _{-1}^{1}(9-6x^{2}+x^{4}-4)\,dx\\&=\pi \int _{-1}^{1}(5-6x^{2}+x^{4})\,dx\\&=\pi \left[5x-2x^{3}+{\frac {1}{5}}x^{5}\right]_{-1}^{1}\\&=\pi \left[\left(5-2+{\frac {1}{5}}\right)-\left(5(-1)-2(-1)^{3}+{\frac {1}{5}}(-1)^{5}\right)\right]\\&=\pi \left(10-4+{\frac {2}{5}}\right)\\&={\frac {32\pi }{5}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/342e24c5799965071cec07df245276cfdf25e6a3)