Science:Math Exam Resources/Courses/MATH101/April 2006/Question 02 (d)/Solution 1

First, we draw a picture. Setting the line and parabola equal to each other, we see that the points of intersection are at

${\begin{aligned}(x-2)^{2}&=x\\x^{2}-4x+4&=x\\x^{2}-5x+4&=0\\(x-4)(x-1)&=0\end{aligned}}$

and so x=1 or x=4. This gives the following picture.

The white area is what we want to rotate. Let x be a point between 1 and 4. Then the radius of a shell is r = 5-x. The height of a shell is $h=x-(x-2)^{2}$. Thus, when we rotate about x=5, the height will move a distance of $2\pi r$ and hence, the volume of revolution is

${\begin{aligned}V&=\int _{1}^{4}2\pi (5-x)(x-(x-2)^{2})\,dx\\&=\int _{1}^{4}2\pi (5-x)(x-x^{2}+4x-4)\,dx\\&=\int _{1}^{4}2\pi (5-x)(-x^{2}+5x-4)\,dx\\&=\int _{1}^{4}2\pi (5(-x^{2}+5x-4)-x(-x^{2}+5x-4))\,dx\\&=\int _{1}^{4}2\pi (-5x^{2}+25x-20+x^{3}-5x^{2}+4x)\,dx\\&=\int _{1}^{4}2\pi (x^{3}-10x^{2}+29x-20)\,dx\\&=2\pi (x^{4}/4-10x^{3}/3+29x^{2}/2-20x){\bigg |}_{1}^{4}\\&=2\pi (4^{4}/4-10(4)^{3}/3+29(4)^{2}/2-20(4)\\&\quad -((1)^{4}/4-10(1)^{3}/3+29(1)^{2}/2-20(1)))\\&=2\pi (64-640/3+29(8)-80-(1/4-10/3+29/2-20))\\&=45\pi /2\end{aligned}}$