Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Solution 1

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

${\begin{aligned}W&=\int _{0}^{3}1000\pi g(2+y)(9-y^{2})dy\\&=1000\pi g\left[\int _{0}^{3}(18+9y-2y^{2}-y^{3})dy\right]\\&=1000\pi g\left[18y+9y^{2}/2-2y^{3}/3-y^{4}/4]_{0}^{3}\right]\\&=1000\pi g\left(18\cdot 3+9(3^{2})/2-2(3^{3})/3-3^{4}/4\right)\end{aligned}}$