Science:Math Exam Resources/Courses/MATH101/April 2018/Question 07 (i)

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MATH101 April 2018

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Question 07 (i)
Let ${\textstyle {\mathcal {R}}}$ be the portion of the circle ${\textstyle (y+1)^{2}+x^{2}=4}$ , with distances measured in metres, that lies to the right of the ${\textstyle y}$ -axis and is below the horizontal line ${\textstyle y=0}$ denoting the ${\textstyle x}$ -axis (see the plot below). A reservoir dug into the ground has the shape that results from the region ${\textstyle {\mathcal {R}}}$ being rotated around the ${\textstyle y}$ -axis. A cross-section of the reservoir is shown below. Area of the reservoir (i) Calculate the volume of the reservoir by slicing horizontally. Simplify your answer completely.

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
Slicing horizontally the reservoir, we get cylindrical pancakes. Find the radius of the pancakes from the equation of the circle and get the volume of the reservoir.

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Solution
Slicing horizontally the reservoir, we get cylindrical pancakes with its height $dy$ . Since the equation for the circle is $(y+1)^{2}+x^{2}=4$ , the piece of the circle which lies to the right of the $y$ -axis can be parametrized by $x={\sqrt {4-(y+1)^{2}}}$ . Therefore, the pancake at the height $y$ from $y=0$ has its radius $x={\sqrt {4-(y+1)^{2}}}$ , and the volume of the pancake at the height $y$ is $dV=\pi \left({\sqrt {4-(y+1)^{2}}}\right)^{2}dy$ Since the range of the height is from $-3$ to $0$ , we finally get the volume formula for the reservoir as ${\begin{aligned}V&=\int _{-3}^{0}\pi (4-(y+1)^{2})dy=\pi \int _{-3}^{0}-y^{2}-2y+3dy\\&=\pi \left[-{\frac {1}{3}}y^{3}-y^{2}+3y\right]_{y=-3}^{y=0}=-\pi \left[-{\frac {1}{3}}(-3)^{3}-(-3)^{2}+3\cdot (-3)\right]\\&=9\pi .\end{aligned}}$ Answer: $\color {blue}9\pi ({\mbox{m}}^{3})$