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A hemispherical barrel of 3 meters of radius is fill for 2 meters of water. I need to calculate the content volume in $m^3$.
I tried this solution: change to polar coordinates, the radius is between 0 and 3, the barrel is hemispherical so $\theta$ is between 0 and $\pi$, and, I supposed that z is the height, so it is between 0 and 2. So:

$\int_0^\pi d\theta\int_0^3\rho d\rho\int_0^2 dz = 9\pi$

but it doesn't work. I know the result that is $\frac{28}3\pi$.
Can you help me and explain where I'm wrong.

1 Answers 1

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Note that $\rho$ depends on $z$ as follows:

enter image description here

So $\rho=\sqrt{(3)^2-(3-z)^2}$.

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    ok, so in this specific case $\rho$ is between 0 ans $2\sqrt{2}$ – 2017-01-08
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    @Mazzorca: No, $\rho$ is between $0$ and $\sqrt{(3)^2-(3-z)^2}$, since $z$ varies between $0$ and $2$. – 2017-01-08