changing an integral from rectangular to spherical

**Frostking** · March 16th 2009, 11:09 AM

I have a triple integral in terms of x,y and z and need to convert it to spherical coordinates.

it is the integral from 0 to 2 and then from 0 to (4 - x^2)^(1/2) inner bounds from 0 to (4 - x^2 - y^2)^(1/2) of (x^2 + y^2 + z^2)^(1/2) dz dy dx

Since the integral is of roe spherical will work well I think!!!

So, we will integrate roe^3 sin phi d roe d phi d theta

My lower bounds must be all zeros.

Can someone please go through the thinking process to determine my upper bounds?

Once I have these I can do the integration no problem. Without them I am stuck! Frostking

**Showcase_22** · March 16th 2009, 11:42 AM

$\int_0^2 \int_0^{\sqrt {4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2} dz dy dx$

$\phi$ varies from 0 to $\pi.$ .
$\theta$ varies from 0 to $2 \pi$ .
$\rho$ varies from 0 to 2.

Hence the integral is:

$\int_0^{2\pi} \int_0^{\pi} \int_0^{2} \rho^3 sin \phi \ d \rho d \phi d \theta$

EDIT: I've changed the $\rho$ limits to how they should be.

**TwistedOne151** · March 16th 2009, 12:04 PM

Shouldn't you have $\rho$ ranging from 0 to 2?

--Kevin C.

**Showcase_22** · March 16th 2009, 12:06 PM

Ah yes, it isn't a unit sphere is it!!

*doi*

**Frostking** · March 16th 2009, 01:36 PM

Yes 0 to 2 pi is what the key has for roe but it has 0 to pi /2 for both the others and I still do not understand how to get any of these??? Any explanation would be very much appreciated. Frostking

**Showcase_22** · March 17th 2009, 03:32 AM

I've only started doing these myself, hence why I made a mistake!

The limits of $\rho$ , $\theta$ and $\phi$ are found from the definition of spherical polar co-ordinates (see the video below):

**Danny** · March 17th 2009, 03:55 AM

Your only dealing with 1/8 of a sphere (first octant only) and in the xy plane the region is a quarter of a circle so the limits should be

$\int_0^{\pi/2} \int_0^{\pi/2} \int_0^{2} \rho^3 sin \phi \ d \rho d \phi d \theta$

**Showcase_22** · March 17th 2009, 04:01 AM

Sorry, but how is it $\frac{1}{4}$ of a circle?

**Danny** · March 17th 2009, 04:48 AM

Originally Posted by Showcase_22

Sorry, but how is it $\frac{1}{4}$ of a circle?

From your earlier post

$\int_0^2 \int_0^{\sqrt {4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2} dz dy dx$ so from your limits

$0 \le x \le 2,\; 0 \le y \le \sqrt{4-x^2}$ , a circle in the first quadrant, i.e. a 1/4 of a circle.

**Showcase_22** · March 17th 2009, 10:01 AM

Oh right, sorry. I think my mind shut down there for a while =S

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