Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06

MATH101 April 2012

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[hide] Question 06
A tank in the shape of a hemispherical bowl of radius 3m, with an outlet that rises 2m above its top (see the diagram below), is full of water. Using the fact that the density of water is $1000{\text{kg/m}}^{3}$ , find the work (in joules) required to pump all the water out of the outlet. You may use the value $g=9.8{\text{m/s}}^{2}$ for the acceleration due to gravity. You do not need to simplify your answer, but you must completely evaluate any integral(s) that arise. Diagram for the above question

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[show] Hint
What is the shape of the water surface and how does it change during the process? To describe this, you need the crucial variable that changes with the process of pumping. You can use either the depth of water or the distance from the top of the bowl to the water level, but the latter will be easier. Then find a formula for the water surface so that you can find the work done to lift up a small slab of water from the water surface up to the top of the bowl and out of the outlet.

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[show] 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. Denote y as the distance from the top of the bowl to the water level and r the radius of the water surface. Then, consider a right-angled triangle with a water surface radius r as base, then the height is y, and the hypoteneuse is 3, which is the radius of the bowl. (The top of the triangle is exactly at the center of the the sphere, half of which becomes the bowl). Then $r^{2}+y^{2}=3^{2}=9$ implies $r^{2}=9-y^{2}$ , so the area of the water surface is given as a function of y by $A(y)=r^{2}\pi =\pi (9-y^{2})$ A thin slab of water thus has volume $A(y)\triangle y$ and mass $M(y)=1000A(y)\triangle y=1000\pi (9-y^{2})\triangle y$ This slab of water travels a distance of y+2 to be pumped out (+2 because of the 2m vertical outlet) and requires a force of Mg Therefore, summing the work done for each slab, and letting $\triangle y\to 0$ , we get ${\begin{aligned}W&=\int _{0}^{3}1000\pi g(2+y)(9-y^{2})dy\\&=1000\pi g\left[\int _{0}^{3}(18+9y-2y^{2}-y^{3})dy\right]\\&=1000\pi g\left[18y+9y^{2}/2-2y^{3}/3-y^{4}/4]_{0}^{3}\right]\\&=1000\pi g\left(18\cdot 3+9(3^{2})/2-2(3^{3})/3-3^{4}/4\right)\end{aligned}}$