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Homework Help: Evaluating a triple integral.

  1. Nov 25, 2009 #1
    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.

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

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

    2. Relevant equations
    Spherical co-ordinates [tex](r,\vartheta,\phi)[/tex] to rectangular co-ordinates [tex](x,y,z)[/tex] are as follows :

    [tex]x=rSin\vartheta Cos\phi[/tex]
    [tex]y=rSin\vartheta Sin\phi[/tex]

    [tex]dV=dxdydz=r^{2}Sin\vartheta dr d\vartheta d\phi[/tex]

    3. The attempt at a solution
    [tex]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[/tex]
    [tex]x^{2}+y^{2}+z^{2} = r^{2}(Sin^{2}\vartheta + Cos^{2}\vartheta)[/tex]
    [tex]x^{2}+y^{2}+z^{2} = r^{2}[/tex]

    But [tex]a^{2}[/tex] is also [tex]x^{2}+y^{2}+z^{2}[/tex]

    So the integral evaluates to

    [tex]\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[/tex]

    Which evaluates to

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

    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 :
    [tex]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[/tex]

    Could somebody explain to me exactly what I need to do to find the boundaries?
  2. jcsd
  3. Nov 25, 2009 #2


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    Your limits are correct except the [itex]\pi[/itex] limit. Since you are limited to the first octant, neither [itex]\theta[/itex] nor [itex]\phi[/itex] can exceed [itex]\pi/2[/itex]. 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 [itex]\theta[/itex] and [itex]\phi[/itex]. Think of moving the radius around the first octant and see that the angles don't get larger than [itex]\pi/2[/itex].
  4. Nov 25, 2009 #3
    Ah yes, that makes sense. It seems simple now.

    [tex]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[/tex]
    is correct, and this finds the volume of one octant?

    This will then evaluate to
    [tex]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)[/tex]

    Which evaluates to:
    [tex]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[/tex]

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

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

    Which evaluates to:
    [tex]e \frac{a^{7}\pi}_{210}}[/tex]

    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 [tex]x = 5a, y = 8, z = 7[/tex], what would be different in how I set the limits?
    Last edited: Nov 25, 2009
  5. Nov 25, 2009 #4


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    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.
  6. Nov 26, 2009 #5
    Sorry, I thought I was calculating the volume of something because I was integrating :
    (expression) dV.

    I read that
    \int\int\int f(x,y,z) dzdydz
    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?
    Last edited: Nov 26, 2009
  7. Nov 26, 2009 #6


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    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?
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