# 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.