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

MATH101 April 2015

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Question 10 (b)
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. (b) Evaluate the integral you wrote down in part (a). (You may use any technique you know, including a geometric one, to do so.)

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 substitution method and the identity $\sin ^{2}x+\cos ^{2}x=1$.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that in part (a), we have the volume of solid $S$ as $V=\int _{-1}^{1}8\pi {\sqrt {1-x^{2}}}dx=8\pi \int _{-1}^{1}{\sqrt {1-x^{2}}}dx=16\pi \int _{0}^{1}{\sqrt {1-x^{2}}}dx.$ The last equality follows from the fact that the integrand is an even function. Now, let's evaluate the integral. We use the substitution $x=\sin t$. This relation implies that $dx=\cos tdt$ and $1-x^{2}=1-\sin ^{2}t=\cos ^{2}t$. Therefore, we have $\int _{0}^{1}{\sqrt {1-x^{2}}}dx=\int _{0}^{\frac {\pi }{2}}{\sqrt {\cos ^{2}t}}\cos tdt=\int _{0}^{\frac {\pi }{2}}\cos ^{2}tdt={\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}1+\cos(2t)dt={\frac {1}{2}}\left[t+{\frac {1}{2}}\sin(2t)\right]_{0}^{\frac {\pi }{2}}={\frac {1}{2}}\cdot {\frac {\pi }{2}}={\frac {\pi }{4}}.$ To sum, we have $V=16\pi \int _{0}^{1}{\sqrt {1-x^{2}}}dx=\color {blue}4\pi ^{2}.$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From part (a), we have the volume of solid $S$ as $V=8\pi \int _{-1}^{1}{\sqrt {1-x^{2}}}dx.$ Let $C_{1}:x^{2}+y^{2}=1$ , be the circle centered at the origin with the radius 1, then for $y>0$, the integrand $y={\sqrt {1-x^{2}}}$ is the semi-circle in the upper half-plane. Therefore, we can interpret the integral $\int _{-1}^{1}{\sqrt {1-x^{2}}}dx$ as the area between the upper part of the circle and x-axis, so that $\int _{-1}^{1}{\sqrt {1-x^{2}}}dx={\frac {1}{2}}\cdot ({\text{area of the circle }}C_{1})={\frac {1}{2}}\pi .$ Plugging this into the volume formula, we have $V=\color {blue}4\pi ^{2}$.