Science:Math Exam Resources/Courses/MATH101/April 2013/Question 04 (b)/Solution 1

To begin with, to get our bounds of integration and where the intersection points are, we follow same steps as in part (a) to find that the curves intersect when $\displaystyle x=\pi /2,\pi ,3\pi /2$. Further, the top function on $\displaystyle [\pi /2,\pi ]$ is $\displaystyle 4+\pi \sin x$, whereas the top function on $\displaystyle [\pi ,3\pi /2]$ is $\displaystyle 4+2\pi -2x$.

Since we are revolving around the horizontal line $\displaystyle y=-1$, the radius of the discs change for every value of x. Hence we will integrate over all x values to get the volume of revolution. To find the radius of the discs as a function of x, note that we need to add 1 to the function value since the radius is given by

$\displaystyle {\text{radius = }}{\text{function}}-{\text{axis}}=f(x)-(-1)=f(x)+1$

where $\displaystyle f(x)$ is the function. The volume of revolution is given by $\displaystyle \pi \int _{a}^{b}({\text{outer radius}}^{2}-{\text{inner radius}}^{2})\ dx$, so

$\displaystyle {\begin{aligned}V&=\pi \int _{\pi /2}^{\pi }[(4+\pi \sin x+1)^{2}-(4+2\pi -2x+1)^{2}]\ dx\\&\qquad +\pi \int _{\pi }^{3\pi /2}[(4+2\pi -2x+1)^{2}-(4+\pi \sin x+1)^{2}]\ dx\end{aligned}}$