# Work to pump water out of a tank with radius $10$

Water density is 1000, tank is a half sphere with radius $10$.

9.8 for gravity, 1000 for density, 2pi to find the volume and all of the rest gives me work.

$$9.8(1000)(2\pi) \int r^2 dy$$

To find the radius I just use $(10 - y)$

This is wrong, but why?

• Is the tank the top half of a sphere or the bottom? That will make a difference here. Commented Jul 7, 2013 at 23:32
• The bottom half. Commented Jul 7, 2013 at 23:34

The mass of an infinitely thin layer of water at depth $y$ is $1000\pi r^2dy$, so the force needed to lift it is $9.8\cdot1000\pi r^2dy$, and the work done in lifting it is $9.8\cdot1000\pi r^2ydy$. Your integral should be
$$9800\pi\int_0^{10}r^2ydy\;.$$
Moreover, $r$ is not $10-y$: $r^2+y^2=10^2$, so $r^2=100-y^2$.
• @Paul: Let $C$ be the point at the centre of the top of the tank. Let $P$ be a point $y$ units directly below $C$, and let $Q$ be any point on the wall of the tank at the same depth as $P$. (Draw a sketch.) Then $\triangle CPQ$ is a right triangle, with right angle at $P$, so we can apply the Pythagorean theorem: $|CP|^2+|PQ|^2=|CQ|^2$. But $|CP|=y$, $|PQ|=r$, and $|CQ|=10$, since the segment $\overline{CQ}$ is a radius of the sphere, so $y^2+r^2=10^2$. Commented Jul 7, 2013 at 23:39