Science:Math Exam Resources/Courses/MATH101/April 2007/Question 01 (e)

MATH101 April 2007

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Question 01 (e)
Calculate the volume of the solid obtained by rotating the region above the x-axis, below the curve $y={\frac {\sin x}{x}}$ and between the lines $x={\frac {\pi }{2}}$ and $\displaystyle x=\pi$ about the y-axis.

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 cylindrical shells. (Shell Integration)

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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. To solve this question, choose an x between $\pi /2$ and $\pi$. From here, the radius of a shell rotated about the y-axis is just r = x and the height of this shell is ${\frac {\sin x}{x}}-0={\frac {\sin x}{x}}$. The distance the height travels aroudn the y-axis is $2\pi r$ and hence, we have ${\begin{aligned}V&=\int _{\pi /2}^{\pi }2\pi (x)(\sin(x)/x)\,dx\\&=\int _{\pi /2}^{\pi }2\pi \sin(x)\,dx\\&=2\pi (-\cos(x)){\bigg |}_{\pi /2}^{\pi }\\&=2\pi (-\cos(\pi )+\cos(\pi /2))\\&=2\pi (-(-1)+0)\\&=2\pi \end{aligned}}$

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