Science:Math Exam Resources/Courses/MATH101/April 2017/Question 07 (a)

MATH101 April 2017

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[hide] Question 07 (a)
Let $R$ be the region to the right of the $y$-axis and to the left of the curve $x^{2}={\sqrt {4+y}}$ with $y$ between $-4$ and $0$. A reservoir dug into the ground has the shape that results from the region $R$ being rotated around the $y$-axis. (a) Calculate the volume of the reservoir (by slicing horizontally). Simplify your answer completely.

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[show] Hint
Draw the region $R$ and the reservoir on $x$-$y$ plane. Then, slice the reservoir horizontally. What's the volume of the slices?

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[show] Solution

The region $R$ and the reservoir are given on $x$-$y$ plane as follows: Slicing the reservoir horizontally, for $y$ between $-4$ and $0$, we have a circle cross section at $y$ whose area is is $\pi x^{2}=\pi {\sqrt {y+4}}$. Combining with the height $dy$ of each slice, we get the volume of the slice at $y$ as $dV=\pi x^{2}dy=\pi {\sqrt {y+4}}dy$. Summing them up for $y\in (-4,0)$, we obtain the volume $V$ of the reservoir; ${\begin{aligned}V=\int _{-4}^{0}dV=\int _{-4}^{0}\pi {\sqrt {y+4}}dy=\pi \int _{0}^{4}{\sqrt {z}}dz={\frac {2}{3}}\pi \left[({\sqrt {z}})^{3}\right]_{0}^{4}={\frac {2}{3}}\pi \cdot 8={\frac {16}{3}}\pi .\end{aligned}}$ Here, the third equality follows from the substitution $z=y+4$. Answer: $\color {blue}{\frac {16}{3}}\pi$