Science:Math Exam Resources/Courses/MATH103/April 2016/Question 05 (a)

MATH103 April 2016

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Question 05 (a)
Let $R$ be the region bounded between the curve $y=-\ln x$ and the lines $y=0,y=1,x=0.$ Find the volume of the solid obtained by revolving $R$ around the y-axis.

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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
Recall that by the disk method, the volume of the solid formed by revolving the region bounded by the y-axis and $f(y)$ about the y-axis itself is $V=\int _{a}^{b}\pi f(y)^{2}\,dy.$

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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. The formula for the volume of the solid obtained by revolving $R$ around the y-axis is $V=\int _{a}^{b}\pi f(y)^{2}\,dy,$ where $f(y)$ is the curve defined by the graph of $y=-\ln(x),$ expressed as a function of $y.$ Solving for $x$ in $y=-\ln(x)$ yields $x=e^{-y}$, i.e., $f(y)=e^{-y}.$ The graph of the region $R$ also shows that $y$ varies between $0$ and $1,$ so $a=0$ and $b=1.$ Hence by the formula ${\begin{aligned}V&=\int _{0}^{1}\pi e^{-2y}\,dy\\&=\left.\pi \cdot {\frac {e^{-2y}}{-2}}\right|_{0}^{1}\\&=-{\frac {\pi }{2}}\left(e^{-2}-1\right)\\&=\color {blue}{\frac {\pi }{2}}(1-e^{-2}).\end{aligned}}$