spherical coordinates!!!

**althaemenes** · March 21st 2009, 10:32 AM

Hi,
For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:

many thanks....

Problem:

Integrate the function

over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
integral =
(You should find the actual value when you evaluate the integral!)

**HallsofIvy** · March 21st 2009, 11:34 AM

Originally Posted by althaemenes

Hi,
For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:

many thanks....

Problem:

Integrate the function

over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
integral =
(You should find the actual value when you evaluate the integral!)

Because of the spherical symmetry, the limits of integration in spherical coordinates are very simple. We want to go completely around so $\theta$ goes from 0 to $2\pi$ . We want to go from the z axis down to the xy-plane so $\phi$ goes from 0 to $\pi/2$ . Finally, of course, $\rho$ goes from 1 to 4.

I presume you know that, in polar coordinates, $x= \rho sin(\phi)cos(\theta)$ and $y= \rho sin(\phi)sin(\theta)$ so $-5x+ 5y= -5\rho sin(\phi)cos(\theta)+ 5 sin(\phi)sin(\theta)$ $= 5sin(\phi)(sin(\theta)- cos(\theta))$

Of course, the differential of volume in spherical coordinates is $\rho^2 sin(\phi)d\rho d\theta d\phi$ .

**Showcase_22** · March 21st 2009, 11:46 AM

$\int_1^4 \int_0^\frac{\pi}{2} \int_0^{2\pi} 5sin (\phi) (sin (\theta)-cos (\theta)) \rho^2 sin(\phi)d \rho d\phi d\theta$

I was having trouble deciding what the $\rho$ limits should be until I read HallsofIvy's post. Thanks!

**althaemenes** · March 21st 2009, 01:55 PM

HallsofIvy and Showcase_22

!!! Thanks guys!!

=> (105/2)*Pi-(105/2)*Pi*sin(Pi)*cos(Pi)-(105/2)*Pi*cos(Pi)^2

=> 0

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