Science:Math Exam Resources/Courses/MATH101/April 2010/Question 02 (a)/Solution 2

We see the function and a sketch of the volume in the following figure.

We use the washers method

$V=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx$

to find the volume, where R_O is the outer radius of the washer and R_I is inner radius. From the diagram above, we see that $R_{O}(x)=1-x^{2}+2=3-x^{2}$, and $R_{I}(x)=2$.

We are integrating along x, so the boundaries of the integral must be $x=[-1,1]$. Now we need to calculate the integral

${\begin{aligned}V&=\pi \int (R_{O}(x)^{2}-R_{I}(x)^{2})\,dx\\&=\pi \int _{-1}^{1}((3-x^{2})^{2}-4)\,dx\\&=\pi \int _{-1}^{1}(9-6x^{2}+x^{4}-4)\,dx\\&=\pi \int _{-1}^{1}(5-6x^{2}+x^{4})\,dx\\&=\pi \left[5x-2x^{3}+{\frac {1}{5}}x^{5}\right]_{-1}^{1}\\&=\pi \left[\left(5-2+{\frac {1}{5}}\right)-\left(5(-1)-2(-1)^{3}+{\frac {1}{5}}(-1)^{5}\right)\right]\\&=\pi \left(10-4+{\frac {2}{5}}\right)\\&={\frac {32\pi }{5}}\end{aligned}}$