A tank full of water has the shape obtained by revolving the curve $y = arcsin(x)$ around the y axis from $x = 0$ to $x = 1$. Find the work required to pump the water out of the tank. (The density of water is $1000$ kg / m$^3$ ).
I know I have to use integration to solve the problem, but I don't know anything other than that.
Show that the work involved in pumping a cross-sectional element of the water is
$$\pi \rho g x^2(y) (1-y) dy$$
where $x(y) = \sin{y}$. $\rho$ is the density and $g$ is acceleration due to gravity. $1-y$ is the distance needed to raise the element to pump it out of the tank. The the total work involved is
$$\pi \rho g \int_0^1 dy \, (1-y) \sin^2{y} $$
This integral may be done using integration by parts. The result is that the work is
$$\frac{\pi}{4} \rho g \cos^2{(1)}$$