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?
Best Answer
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 $r^2=100-y^2$.