Science:Math Exam Resources/Courses/MATH101/April 2009/Question 03 (b)/Solution 1

A picture is included below. To solve this question, we need to use the method of left and right side integration. Recall the formula

$V=\pi \int _{a}^{b}((x_{R}+s)^{2}-(x_{L}+s)^{2})\,dy$

where we think of $x_{R}$ and $x_{L}$ as functions of y and the $s$ factor is the shifting factor (so basically this factor accounts for what line we are rotating over - the picture should make this clear). Here, the right most curve is $x_{R}=\arccos(y)$ and the left most curve is $x_{L}=0$ and the $s$ factor in this case is 2. The last thing is the limits of integration. we know on the x-axis, from the previous problem, we were integrating from 0 to ${\tfrac {\pi }{2}}$. Here however we are integrating from the y-axis so we want to know when $\cos(x)$ crosses the y-axis. This occurs when $y=1$. Thus, we have

$V=\pi \int _{0}^{1}((\arccos(y)+2)^{2}-(2)^{2})\,dy=\pi \int _{0}^{1}((\arccos(y))^{2}+4\arccos(y))\,dy$

To challenge yourself, try showing the above is actually equal to $\displaystyle \pi ^{2}+2\pi$.