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find the center of mass of the upper half of the ball $x^2+y^2+z^2 \le 16$ for $z\ge 0$

can someone help me set up the integrals?

Its symmetry on the x and y axis right so we only have to find z-axis.

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This describes a hemispherical volume. Perhaps you can find a simple set of limits using spherical coordinates? – Muphrid Nov 19 '12 at 23:41
i hate spherical coordinates ehh what would the limits be though 0 to 2pi on phi and 0 to 2pi on delta what about p? 0 to 4? – Jack F Nov 19 '12 at 23:46
up vote 2 down vote accepted

Imagine the half-ball is a ham, and take a thin horizontal slice going from height $z$ to height $z+dz$. The cross section at height $z$ is a disk. Since $x^2+y^2=16-z^2$, the disk has radius $\sqrt{16-z^2}$. This disk has area $\pi(16-z^2)$. So the volume of the slice is about $\pi(16-z^2)\,dz$.

This slice is at height $z$ above the $x$-$y$ plane, so it has moment approximately equal to $$(z)(\pi)(16-z^2)\,dz$$ about the $x$-$y$ plane.

"Add up," $z=0$ to $z=4$. We get that the moment is $$\int_0^4\pi(16z-z^3)\,dz.$$ Evaluate this integral, divide by the volume of the half-ball.

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