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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?

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    $\begingroup$ Is the tank the top half of a sphere or the bottom? That will make a difference here. $\endgroup$
    – Nick Peterson
    Commented Jul 7, 2013 at 23:32
  • $\begingroup$ The bottom half. $\endgroup$
    – Paul the Pirate
    Commented Jul 7, 2013 at 23:34

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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 $r^2=100-y^2$.

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  • $\begingroup$ I know I can just memorize that pattern but why is the radius that? $\endgroup$
    – Paul the Pirate
    Commented Jul 7, 2013 at 23:35
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    $\begingroup$ @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 $y^2+r^2=10^2$. $\endgroup$
    – Brian M. Scott
    Commented Jul 7, 2013 at 23:39

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