Calculus 3 question

[Guest]

Hi everyone
How are you?

I'm having trouble with this question here:

Find the moment Mxy with respect to the xy-plane for the top half of the unit ball centered at the origin. That is, evaluate

SSSR z dV,
where R is the portion of the unit ball centered at the origin that lies about the xy-plane.

p.s.
The S's are the triple integral signs.

I know how to do a unit ball if it's the whole unit ball, but I don't know what my limits would be if I only take the top half.

This is what I'm assuming it to be:

c <= pi/2 <= d, weird symbol <= pi <= weird symbol, 0 <= p <= 1 and then z would be cos phi, I think phi is the right word for it, since z = cos phi

Anyway, thanks for the help on this.

Take care,
Beckie

 

[galactus]

In spherical coordinates, \(\displaystyle z={\rho}cos({\phi})\)

In rectangular coordinates we have:

\(\displaystyle \int_{-\1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_0^{\sqrt{1-x^{2}-y^{2}}zdzdydx\)

In spherical coordinates we have:

\(\displaystyle \int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^{1}{\rho}cos({\phi}){\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}\)

Is this what you wanted?. I hope I'm correct after all that LaTexing. :lol:

FYI, I don't know what your "weird symbol" is but that "p-looking" symbol is the Greek letter rho, the little circle with the line through it is phi, and of course, you know theta.

 

[Guest]

Hi Galactus
Thank you thank you thank you thank you!!

You are my hero

I wish I knew how to put those little symbols in.

How come theta is 2pi and 0? Is it because if I take the z^2 out of the equation it would become x^2 + y^2 = 1 and then that limit would be 2pi and 0?

Does the pi/2 come from the x^2 + y^2 + z^2 = 1 with the top half of the sphere?

Sorry if I'm not wording this right. Let me know if this sounds confusing.

Thanks
Take care,
Beckie

 

[galactus]

Think of \(\displaystyle {\phi}\) as latitude and \(\displaystyle {\theta}\) as longitude.

Picture \(\displaystyle {\phi}\) running from the North Pole down to the equator. Because you only have half a sphere, you only need \(\displaystyle \frac{\pi}{2}\).

\(\displaystyle {\theta}\) goes around the equator. It goes from 0 to \(\displaystyle 2{\pi}\)

because it completes the circle in the xy plane.

Clear as mud?. There should be a picture in your text explaining this.

If you're interested in knowing how I formatted the post, click on quote and check out the LaTex code I used.