From Cartesian to spherical integral

[Feodalherren]

Homework Statement

Homework Equations

[itex][/itex]

The Attempt at a Solution

Is this the correct setup?

$\int^{\pi}_{\frac{3\pi}{4}}\int^{2\pi}_{0}\int^{\sqrt{2}}_{0}\frac{1}{\rho^{2}} rho^2 Sin\phi d\rho d\theta d\phi$

I gave up on itex. It was either that or my computer flying out the window. My apologies.

 

[Simon Bridge]

You came up with:
$$\int_{3\pi/4}^\pi \int_0^{2\pi}\int_0^{\sqrt{2}}\frac{\sin\phi\;\text{d}\rho \text{d} \theta \text{d}\phi}
{\rho^2}$$
(use the quote button on this post to see how I did that ;) )

Please show your working.

 

[Feodalherren]

God, that looks complicated. I'll dedicate some time to actually learning how to write those integrals as nicely as you do after this week. I'm drowning in chemistry and math right now and haven't got a second to spare to learn itex right now :).

What I did was I drew a picture and then basically looked at it.
The limits for theta are obvious. The phi I got from setting x=0 on the upper limit and getting z=-x in the zy plane, this is 45 degree line, it's negative so 3pi / 4.

As for the rho limits, that's where I get less sure of what I'm doing.

I think that what I'm looking for is an ice cream cone-shape. Therefore the rho starts from 0 and ... Oh no wait I think I see it now. If I set the z-limits equal to each other I get the intercept which is r=1.
So then rho would go from 0 to 1. Correct?

 

[LCKurtz]

Homework Statement

Homework Equations

[itex][/itex]

The Attempt at a Solution

Is this the correct setup?

$\int^{\pi}_{\frac{3\pi}{4}}\int^{2\pi}_{0}\int^{\sqrt{2}}_{0}\frac{1}{\rho^{2}} rho^2 Sin\phi d\rho d\theta d\phi$

I gave up on itex. It was either that or my computer flying out the window. My apologies.

That is correct except why not cancel the $\rho^2$ factors?

As for the rho limits, that's where I get less sure of what I'm doing.

I think that what I'm looking for is an ice cream cone-shape. Therefore the rho starts from 0 and ... Oh no wait I think I see it now. If I set the z-limits equal to each other I get the intercept which is r=1.
So then rho would go from 0 to 1. Correct?

No, $\rho$ isn't the same as $r$. Your original limits are correct.