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

MATH101 April 2017

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Question 07 (b)

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. (b) The reservoir from part (a) is filled with a fluid of density $1234{\text{kg}}/{\text{m}}^{3}$. Find the work, in joules, required to pump the fluid out of the top of the reservoir. Use $g=9.8{\text{m}}/{\text{s}}^{2}$ for the acceleration due to gravity. A calculator-ready answer is sufficient.

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
Consider the slices from the part (a) again. How much is the work required to pump the fluid in the slice at $y$ out of the top of the reservoir?

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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. Recall that in Physics, the work $W$ that a constant force $F$ does when moving an object over a distance of $d$ is $W=Fd$. On the other hand, by the Newton's Second Law, the force made by gravity is given as $F=mg=\rho Vg$, where $\rho$ and $V$ are the density and the volume of the object. Now, we consider the slices from the part (a) again and find the work required to pump the fluid in the slice at $y$ out of the top of the reservoir. In part (a), we find the volume of the slice at $y$ as $dV=\pi {\sqrt {y+4}}dy$. On the other hand, the distance between the fluid in the slice at $y$ and the top of the reservoir is obviously $|y|=-y$. Combining with the constant density $1234{\text{kg}}/{\text{m}}^{3}$ and the acceleration due to gravity $g=9.8{\text{m}}/{\text{s}}^{2}$, the work for the slice at $y$ is $dW=\rho \cdot dV\cdot g\cdot d=(1234)\cdot (9.8)\cdot \pi (-y){\sqrt {y+4}}dy$. Summing them up for $y\in (-4,0)$ and using the same substitution $z=y+4$ from part (a), we can find the desired work as follows: ${\begin{aligned}W&=-\int _{-4}^{0}(1234)\cdot (9.8)\cdot \pi y{\sqrt {y+4}}dy=-(1234)\cdot (9.8)\pi \int _{0}^{4}(z-4){\sqrt {z}}dz\\&=-(1234)\cdot (9.8)\pi \left(\int _{0}^{4}{\sqrt {z}}^{3}dz-\int _{0}^{4}4{\sqrt {z}}dz\right)\\&=-(1234)\cdot (9.8)\pi \left(\left[{\frac {2}{5}}({\sqrt {z}})^{5}\right]_{0}^{4}-4\cdot \left[{\frac {2}{3}}({\sqrt {z}})^{3}\right]_{0}^{4}\right)\\&=-(1234)\cdot (9.8)\pi \left({\frac {64}{5}}-{\frac {64}{3}}\right)=(1234)\cdot (9.8)\pi \cdot 64\cdot {\frac {2}{15}}\end{aligned}}$ Considering units, the answer is $\color {blue}(1234)\cdot (9.8)\pi \cdot 64\cdot {\frac {2}{15}}J$