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

$\begin{array}{rl}V& =\pi \int (R_O(x{)}^2-R_I(x{)}^2)dx\\ & =\pi \underset{-1}{\overset{1}{\int }}((3-x^2{)}^2-4)dx\\ & =\pi \underset{-1}{\overset{1}{\int }}(9-6x^2+x^4-4)dx\\ & =\pi \underset{-1}{\overset{1}{\int }}(5-6x^2+x^4)dx\\ & =\pi {[5x-2x^3+\frac{1}{5}{x}^5]}_{-1}^1\\ & =\pi [(5-2+\frac{1}{5})-(5(-1)-2(-1{)}^3+\frac{1}{5}(-1{)}^5)]\\ & =\pi (10-4+\frac{2}{5})\\ & =\frac{32\pi }{5}\end{array}$