Science:Math Exam Resources/Courses/MATH101/April 2018/Question 03 (ii)

MATH101 April 2018

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Question 03 (ii)
Determine explicitly the volume of revolution obtained by rotating the bounded region between ${\textstyle y=x^{2}}$ and ${\textstyle y=2-x^{2}}$ about the ${\textstyle x}$ -axis.

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Hint
Suppose that two functions $f$ and $g$ intersect at $x=a$ and $x=b$ , and satisfy $f(x)\geq g(x)$ on the interval $(a,b)$ . Then, the volume of revolution obtained by rotating the bounded region between ${\textstyle y=f(x)}$ and ${\textstyle y=g(x)}$ about the ${\textstyle x}$ -axis is $\pi \int _{a}^{b}(f(x)^{2}-g(x)^{2})dx.$

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, we find the intersection points between two curves. It is enough to solve $x^{2}=2-x^{2}\iff 2x^{2}=2\iff x^{2}=1\iff x=\pm 1.$ Then, we can observe that $f(x)=2-x^{2}\geq x^{2}=g(x)$ on $(-1,1)$ . (This can be done either by drawing the graphs of the two functions or by comparing the value of the two function at some point in the interval.) Note that since both functions $f$ and $g$ are even, the function $y=f(x)^{2}-g(x)^{2}$ is also even. Thus we have the volume ${\begin{aligned}\pi \int _{-1}^{1}(f(x)^{2}-g(x)^{2})dx&=2\pi \int _{0}^{1}(f(x)^{2}-g(x)^{2})dx\\&=2\pi \int _{0}^{1}((2-x^{2})^{2}-(x^{2})^{2})dx\\&=2\pi \int _{0}^{1}(4-4x^{2})dx\\&=2\pi \left[4x-{\frac {4}{3}}x^{3}\right]_{x=0}^{x=1}\\&=2\pi \left(4-{\frac {4}{3}}\right)\\&=2\pi \cdot {\frac {8}{3}}\\&={\frac {16}{3}}\pi .\end{aligned}}$ Answer: $\color {blue}{\frac {16}{3}}\pi$