Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)/Solution 1

We use the method of washers.

For any x in (-1,2), the larger and smaller radii R and r of the washer is given by

$R(x)=5-(x+1)=4-x\,$

$r(x)=5-(4-(x-1)^{2})=1+(x-1)^{2}\,$

(The simplifications given above are not really necessary)

Notice that the larger radius R is given by the distance from the axis of revolution y=5 to the lower curve, because it is further away.

Therefore, the area of the washer is

$A(x)=(R^{2}-r^{2})\pi =\left[(4-x)^{2}-\left(1+(x-1)^{2}\right)^{2}\right]\pi$

Now we can write down the integral that represents the volume of the solid of revolution:

${\begin{aligned}V&=\int _{-1}^{2}A(x)dx\\&=\pi \left[\int _{-1}^{2}(4-x)^{2}-\left(1+(x-1)^{2}\right)^{2}dx\right]\\\end{aligned}}$

According to the question statement, we may stop here.

If you are interested to find the value of the integral, read on.

To solve this integral, we split it in two parts and write

$V=\pi \left[\int _{-1}^{2}(4-x)^{2}dx-\int _{-1}^{2}\left(1+(x-1)^{2}\right)^{2}dx\right]$

We refrain from expanding the brackets but make a change of variables u=4-x, du=-dx and v=x-1, dv=dx for the first and second integral respectively. Not forgetting to update the integration limits, we arrive at

${\begin{aligned}V&=\pi \left[\int _{-1}^{2}(4-x)^{2}dx-\int _{-1}^{2}\left(1+(x-1)^{2}\right)^{2}dx\right]\\&=\pi \left[\int _{5}^{2}u^{2}(-du)-\int _{-2}^{1}\left(1+v^{2}\right)^{2}dv\right]\\&=\pi \left[\int _{2}^{5}u^{2}du-\int _{-2}^{1}\left(1+2v^{2}+v^{4}\right)dv\right]\\&=\pi \left(\left[u^{3}/3\right]_{2}^{5}-\left[v+2v^{3}/3+v^{5}/5\right]_{-2}^{1}\right)\\&=\pi \left((125-8)/3-\left[3+2(1+8)/3+(1+32)/5\right]\right)\\&=\pi \left(39-3-6-6-3/5\right)\\&=\pi \left(24-3/5\right)=117\pi /5\\\end{aligned}}$