How Do You Convert a Triple Integral Into Spherical Coordinates?

[KStolen]

I'm taking a Calculus class as an elective. This might not have been a good idea, but I'm stuck in it now.
Here is a problem I have to do. My knowledge of basic maths is poor, so please be gentle and explain thoroughly!

1. The problem.
Rewrite the following integral in terms of spherical polar coordinates.

\int\int\int x^{2}z^{2}exp((x^{2}+y^{2}+z^{2})/a^{2})dxdydz

Now evaluate it over the region bounded by the planes x = 0, y = 0, z = 0
and the sphere x^{2} +y^{2} +z^{2} = a^{2} in the first octant.

Homework Equations

Spherical co-ordinates (r,\vartheta,\phi) to rectangular co-ordinates (x,y,z) are as follows :

x=rSin\vartheta Cos\phi
y=rSin\vartheta Sin\phi
z=rCos\vartheta

dV=dxdydz=r^{2}Sin\vartheta dr d\vartheta d\phi

The Attempt at a Solution

x^{2}+y^{2}+z^{2} = r^{2}Sin^{2}\vartheta Cos^{2}\phi + r^{2}Sin^{2}\vartheta Sin^{2}\phi + r^{2}Cos^{2}\vartheta
x^{2}+y^{2}+z^{2} = r^{2}(Sin^{2}\vartheta + Cos^{2}\vartheta)
x^{2}+y^{2}+z^{2} = r^{2}

But a^{2} is also x^{2}+y^{2}+z^{2}

So the integral evaluates to

\int\int\int r^{2}Sin^{2} \vartheta Cos^{2}\phi r^{2}Cos^{2}\vartheta exp(a^{2}/a^{2})r^{2}Sin\vartheta dr d\vartheta d\phi

Which evaluates to

e\int\int\int r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi

So I've written it in spherical co-ordinates. I think everything is right so far. (hopefully I don't have any typos in my LaTeX code)
The next part is what I'm unsure about. I don't know how to set up the boundaries.
No doubt this is trivial, but it's also a central part of doing the integral.

Here's an attempt :
e\int^{\pi}_{0}\int^{\pi/2}_{0}\int^{a}_{0} r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi

Could somebody explain to me exactly what I need to do to find the boundaries?
Thanks!

 

[LCKurtz]

Your limits are correct except the \pi limit. Since you are limited to the first octant, neither \theta nor \phi can exceed \pi/2. It's best to draw a picture to visualize it. Draw the first octant portion of the sphere and a radius from the origin to the boundary of the sphere somewhere in the middle. Label the two variables \theta and \phi. Think of moving the radius around the first octant and see that the angles don't get larger than \pi/2.

 

[KStolen]

Ah yes, that makes sense. It seems simple now.

So
e\int^{\pi/2}_{0}\int^{\pi/2}_{0}\int^{a}_{0} r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi
is correct, and this finds the volume of one octant?

This will then evaluate to
e\int^{\pi/2}_{0}\int^{\pi/2}_{0} a^{7}/7 (Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi d\vartheta d\phi)

Which evaluates to:
e \frac{a^{7}}_{7}}\int^{\pi/2}_{0}\ ((\frac{-Cos\vartheta}{8}-\frac{Cos3\vartheta}{48}+\frac{Cos5\vartheta}{80})Cos^{2}\phi)|^{\pi/2}_{0}d\phi

Which evaluates to:
e \frac{2a^{7}}_{105}}\int^{\pi/2}_{0} Cos^{2}\phi}d\phi

Which evaluates to:
e \frac{2a^{7}}_{105}} (\frac{\phi/2}{2}+\frac{Sin2\phi}{4})|^{\pi/2}_{0}

Which evaluates to:
e \frac{a^{7}\pi}_{210}}

So this is the volume of the octant? It seems a bit of a peculiar number. Did I make a mistake somewhere?

A question : If I had to evaluate this integral against the same sphere bounded by x = 5a, y = 8, z = 7, what would be different in how I set the limits?

 

[LCKurtz]

Which evaluates to:
e \frac{a^{7}\pi}_{210}}

So this is the volume of the octant? It seems a bit of a peculiar number. Did I make a mistake somewhere?

A question : If I had to evaluate this integral against the same sphere bounded by x = 5a, y = 8, z = 7, what would be different in how I set the limits?

No, that is not the volume. You weren't calculating the volume or you would have been integrating 1 dV instead. I didn't check your integrals but, in any case, there is no way to know whether that answer is "reasonable" or not since it doesn't represent volume anyway.

Generally you need to look at the graph to figure out the limits. Your sphere might not even intersect the planes you suggest in your last question if a isn't large enough.

 

[KStolen]

Sorry, I thought I was calculating the volume of something because I was integrating :
(expression) dV.

I read that
<br /> \int\int\int f(x,y,z) dzdydz<br />
Represents the mass of an object, if f(x,y,z) is the density at (x,y,z).

So I see if you integrated 1 dV, you'd get the volume.

But if you have an expression like the one I had to integrate, does it not necessarily mean anything? Would I have to graph/plot it to see what it is?

 

[LCKurtz]

Sorry, I thought I was calculating the volume of something because I was integrating :
(expression) dV.

I read that
<br /> \int\int\int f(x,y,z) dzdydz<br />
Represents the mass of an object, if f(x,y,z) is the density at (x,y,z).

So I see if you integrated 1 dV, you'd get the volume.

But if you have an expression like the one I had to integrate, does it not necessarily mean anything? Would I have to graph/plot it to see what it is?

A stand-alone integral like that in a calculus book may not have any particular meaning; it is just a practice problem. You are correct that if f(x,y,z) ≥ 0 it may represent a density, or maybe not. You may have terms like x² or x²+y² in the integral also if you are doing moments. Without any context, don't worry about what it means. It's just an integral. And without context, plotting it wouldn't help. How would you plot a 3d density function?