Question

) A scientific instrument is encased in a diving bell which has a shape that is roughly approximated by rotating the region R around the y-axis, where R is the region under the curve y arccos(x) (in meters) over the interval [0, 1]. On its most recent trip underwater, the diving bell sprung a small leak at the top and filled up with water at a depth that was half its total height. The water has constant density a in units of ka/m, and must be pumped out of the diving bell through a small pipe 1/4 m long that is added to the topi. U set up an integral to find the work done to lift the water out of the diving bell through the pipe. Calculate the value of this integral. (Note: Do you need to account for gravity? If so, use981m/sec It's fine to simply leave your answer in terms of the constants δ and g, as needed, but do evaluate your integral otherwise.) the result looks a little like a Hershey's kiss where the pipe is a tag sticking straight up out the top!

Answer 1

The water is filled up to $\pi/4$ , The gravitational potential energy of the water (assuming the base at 0 potential)

$U_1=\int \delta gy\pi x^2dy$

$U_1=\int_0^{\pi/4} \delta gy\pi \cos^2ydy$

$U_1=\delta g\pi\int_0^{\pi/4} y\cos^2ydy$

$U_1=\delta g\pi\left ( \frac{\pi^2}{64}+\frac{\pi}{16}-\frac{1}{8} \right )$

The workdone required is the change in potential energy in raising this water to a height of 1/4 m. So the final PE is

$U_2=\left (\int \delta \pi x^2dy \right )g\left ( \frac{\pi}{2}+\frac{1}{4} \right )$

$U_2=\delta\pi g\left ( \frac{\pi}{2}+\frac{1}{4} \right )\left ( \frac{\pi}{8}+\frac{1}{4} \right )$

Thus,

$W=U_2-U_1=\frac{3\delta g\pi}{64}(\pi^2+2\pi+4)$