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

MATH101 April 2012

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Question 02 (b)
Let R be the bounded region that lies between the curve $y=4-(x-1)^{2}$ and the line $y=x+1$. Write down a definite integral giving the volume of the region obtained by rotating R about the line $y=5$. Do not evaluate this integral.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Use the method of washers (cross-sections). Look at the sketch at part a), the cross-sections are formed by intersecting the region R by a straight line x=c where -1<c<2, and then rotating the line segment about the axis y=5. The result should look like a washer. To find the volume of revolution, find the area of this washer as a function of x and then integrate from the left to right endpoint of the region.

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Solution

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

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