Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (b)/Solution 1

We would use disc integration to evaluate this volume. First, we need to determine the $x$ values that define the endpoints of the interval that we need to integrate over. The endpoints are where the curves intersect, so we solve $4-x^{2}=2-x$ or

$x^{2}-x-2=0\quad \rightarrow \quad (x-2)(x+1)=0$

Thus, the endpoints of the interval $x=-1$ and $x=2$. Since we are using the disc method, we recall that the volume of a single disc with radius $r$ and thickness $\Delta x$ is $\pi r^{2}\Delta x$. Since we are rotating each curve about the $x$-axis, we can treat the value of the function as the height of each disc.

By taking a sample point in the interval $(-1,2)$, we can see that $4-x^{2}>2-x$ in the domain of integration so we can get the volume of the defined solid by taking the volume of the solid with radius $4-x^{2}$ and subtracting the volume with radius $2-x$.

${\begin{aligned}V&=\int _{-1}^{2}\pi (4-x^{2})^{2}\,dx-\int _{-1}^{2}\pi (2-x)^{2}\,dx\\&=\pi \int _{-1}^{2}(4-x^{2})^{2}-(2-x)^{2}\,dx\end{aligned}}$

(Note: If you wish to evaluate the integral, the volume of the solid obtained by rotating the region $R$ about the $x$-axis is $V=108\pi /5$.)