Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.