Write a triple integral in spherical coordinates

1. Apr 24, 2014

Jtechguy21

1. The problem statement, all variables and given/known data

Write a triple integral in spherical coordinates that represents the volume of the part of the sphere
X^2+Y^2+Z^2=16 that lies in the first octant(where x,y, and z are coordinates are all greater than or equal to zero)

2. Relevant equations

So i know this is in rectangular form (x,y,z) trying to get it into (p,Θ,ø)

3. The attempt at a solution

X^2+y^2+z^2=16
p^2=16
p=4

To get Θ the formula is
arccos z/(square root of x^2+y^2+z^2)

when i solve for z I get z=√(16-r^2)

These are the limits i know
The limits for Dz are from z=0 to z=√(16-r^2)
The limits for Dr are from r=0 to r= pi/2

I do not know how to find the limits for DΘ (theta)

since the arccos √(16-r^2)/(square root of x^2+y^2+z^2)
should give me my theta. Θ
but i have no real values for x y and z. so i dont know how to approach this.
thank you

∫ ∫ ∫
rDzDrDΘ
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 24, 2014

SteamKing

Staff Emeritus
You don't seem to understand the problem. You should figure out how a differential element of volume dV is represented in spherical coordinates. In cartesian coordinates, dV = dx dy dz. You appear to have used cylindrical coordinates, r, z, and Θ.

http://en.wikipedia.org/wiki/Spherical_coordinate_system

3. Apr 24, 2014

Zondrina

Indeed, and in applying that transformation properly, you will wind up needing to solve:

$r sin(\theta) cos(\phi) ≥ 0$
$r sin(\theta) sin(\phi) ≥ 0$
$r cos(\theta) ≥ 0$

Don't forget the Jacobian.

4. Apr 24, 2014

LCKurtz

@JtechGuy21: Don't set that problem up in rectangular coordinates and try to change it to spherical coordinates. You don't need to change variables and mess with Jacobians. Just remember$$\text{Volume} =\iiint_V 1~dV$$All you need is the $dV$ formula in spherical coordinates and then to put the $\rho,\phi,\theta$ limits on. They are very easy and can be easily seen geometrically.

Last edited: Apr 24, 2014
5. Apr 25, 2014

Jtechguy21

aaaaaah okay.
yes i was turning them into cylindrical for some reason...

Thanks for pointing that out.

p=4
x=4SinøCosΘ
y=4SinøSinΘ
Z=4Cosø

1Dv=p^2sin dp dø dΘ

p^2 Sinø dp dø dΘ

So that means my limits for;
p is 0 to 4
Θ is 0 to pi/2 (because its quadrant one)
and i have no idea how to find my ø limits?
Since i dont have a real value for z. I cant take the arc cos of z to get my ø?

6. Apr 25, 2014

LCKurtz

It would be good for you to quote who you are replying to.

Look at the picture of the sphere in the first octant. Draw a radius and label $\rho$, $\phi$ and $\theta$. It is obvious from the picture what their ranges are. You have $\rho$ and $\theta$ correct so far.

7. Apr 25, 2014

Jtechguy21

hello.
i dont think i need to use a jacobian here. im in the section that involves using triple integrals in cylindrical and spherical coordinates.(not saying its not possible )

The section afterwards is change of variables:jacobians.
Which I've done before, but since we are finding volume in this case, i dont know how jacobians would help since jacobians deal with area in the uv plane in comparison to the xy plane

8. Apr 25, 2014

Jtechguy21

This might be a dumb question, but how do i draw this picture so i could visually see whats happening.
Do i have to put it into z= format?
and use wolfgram alpha

9. Apr 25, 2014

LCKurtz

Just draw it by hand. Draw a section of a sphere that is in the first octant. Surely your book has such pictures. Probably right where it introduces spherical coordinates.

Or look here: http://mathinsight.org/spherical_coordinates

It isn't hard to see the limits for the variables in the first octant. Be aware that physics and math books have the $\phi$ and $\theta$ switched. That link uses the math convention.