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

MATH101 April 2012

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q3 (d) • Q4 (a) • Q4 (c) • Q5 • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) •

Other MATH101 Exams

• April 2011 • April 2010 • April 2009 • April 2008 • April 2007 • April 2005 • April 2006 • April 2012 • April 2013 • April 2014 • April 2015 • April 2016 • April 2018 • April 2017 •

[hide] 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!

[show] 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.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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