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

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MATH103 April 2016

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) (i) • Q6 (a) (ii) • Q6 (a) (iii) • Q6 (a) (iv) • Q6 (a) (v) • Q6 (b) (i) • Q6 (b) (ii) • Q6 (b) (iii) • Q7 (a) (i) • Q7 (a) (ii) • Q7 (b) (i) • Q7 (b) (ii) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) •

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Question 05 (b)
The solid of part (i) is filled with liquid of varying density. It is given that liquid density $y$ units from the bottom of the solid is $e^{-y}$ grams per cubic unit. Find the mass of the liquid inside the solid.

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
Denoting the solid by $S,$ for a given density function $\rho (x,y,z),$ we have the mass formula $m_{S}=\int _{S}\rho \,dV=\int _{S}\rho (x,y,z)\,dx\,dy\,dz.$

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution
The density function of the liquid is given in the question as $\rho (x,y,z)=e^{-y}.$ By the hint, the mass $m_{S}$ of the solid is $m_{S}=\int _{S}e^{-y}\,dV.$ We consider slices of the solid $S$ made at heights $y$ above the x-axis, where each slice has thickness $dy$ and cross-sectional area $A(y).$ In the case of our solid, the cross-sections at height $y$ are circles with radii $e^{-y}$ , so that $A(y)=\pi e^{-2y}.$ Therefore $dV=A(y)\,dy=\pi e^{-2y}\,dy$ (for $0\leq y\leq 1$ ). Returning to the mass formula, this implies that ${\begin{aligned}m_{S}&=\int _{S}e^{-y}\,dV\\&=\int _{0}^{1}e^{-y}\cdot \pi e^{-2y}\,dy\\&=\pi \int _{0}^{1}e^{-3y}\,dy\\&=-{\frac {\pi }{3}}\cdot \left.e^{-3y}\right\|_{y=0}^{y=1}\\&={\frac {\pi }{3}}(1-e^{-3}).\end{aligned}}$ Hence the mass of the liquid inside the solid is $\color {blue}{\frac {\pi }{3}}(1-e^{-3})\,\mathrm {g} .$