Re: Integrals In Spherical Coordinates

From: Daniel Grubb (grubb_at_lola.math.niu.edu)
Date: 11/22/04 

Date: 22 Nov 2004 16:25:39 GMT

>This is homework so I don't want the answer....just some hints.

>Let W be the solid bounded above by x^2+y^2+z^2=9 and below by phi=pi/3.
>Calculate the mass of W if the density at each point is directly
>proportional to the distance above the xy plane.

>As I have it drawn, it is an upward cone about the origin with a slanted
>plane of z = -x - y +3 for a top. I am having a little difficulty setting
>up the triple integral as well as the density function. Here is what I have
>so far:

First of all, you should go back and do some algebra. I suspect that
you solved for z by going from z^2=9-x^2-y^2 and taking square roots
to get z=3-x-y. If that is the case, I would almost immediately give
you 0 points for this problem. Nobody in third semester claculus
should do this mistake *ever*.

>2 pi pi/6 ???
> / / /
> | | | ??? p^2 sin phi d-rho d-phi d-theta
>/ / /
>0 0 0

>Anyone got any pointers on this one?

Well, your order is messed up. To do any multiple integral,
figure out the inside limits *first*. The reasoning goes as follows:

If phi and theta are fixed, what are the allowed values of rho
for the solid? You can think of the rho values as sweeping out
a little radial line. These rho values can depend on both phi and theta.
The smallest value is the lower limit on the inside integral, and the
largest the upper limit.

Now, let phi vary but keep theta fixed. As phi varies, the little radial
line sweeps out a planar cross-section of your volume. What
range of phi values are allowed in your figure? These values may depend
on theta *but not rho* and give the limits of the middle integral.

Finally, what range of theta values are needed for the planar cross
sections to sweep out the whole volume? These values won't depend on
either rho or phi and give the limits for the outside integral.

--Dan Grubb

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