Science:Math Exam Resources/Courses/MATH101/April 2015/Question 10 (a)

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MATH101 April 2015

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Question 10 (a)
Let $R$ be the region inside the circle $x^{2}+(y-2)^{2}=1$ . Let S be the solid obtained by rotating R about the x-axis. (a) Write down an integral representing the volume of S.

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
Sketch the circle, take a slice of that, and revolve it over the x-axis. You should get a thin doughnut, what are the height and inner/outer radius of this doughnut?

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Solution
As we mentioned in Hint, first we sketch the circle. Note that at each $x$ between -1 and 1, the corresponding point in the upper part of circle (above $y=2$ ) is $(x,2+{\sqrt {1-x^{2}}})$ , while the point in the lower part is $(x,2-{\sqrt {1-x^{2}}})$ . Then, take a vertical slice at $x$ with the width $dx$ and rotate it with respect to the x-axis. Then, we get a thin doughnut with the height $dx$ , outer radius $2+{\sqrt {1-x^{2}}}$ , and inner radius $2-{\sqrt {1-x^{2}}}$ . Therefore, the volume $V_{x}$ of this doughnut is ${\begin{aligned}V_{x}&={\text{( volume of outer cylinder)}}-{\text{(volume of inner cylinder)}}=\pi ((2+{\sqrt {1-x^{2}}})^{2}-(2-{\sqrt {1-x^{2}}})^{2})dx\\&=\pi ((2+{\sqrt {1-x^{2}}})-(2-{\sqrt {1-x^{2}}}))((2+{\sqrt {1-x^{2}}})+(2-{\sqrt {1-x^{2}}}))dx=8\pi {\sqrt {1-x^{2}}}dx\end{aligned}}$ Finally, since the volume of the solid $S$ is obtained by integrating $V_{x}$ on $[-1,1]$ , we have $V=\int _{-1}^{1}V_{x}dx=\color {blue}\int _{-1}^{1}8\pi {\sqrt {1-x^{2}}}dx.$