Triple Integral

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  • #1
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Hi everyone. I am trying to integrate the following:

[itex]\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ[/itex]

Here's my work:

[itex]=2\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}r\sqrt{a^{2}-r^{2}}drdθ[/itex]

I use substitution with u=a2-r2 to get:

[itex]=-\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{a^{2}sin^{2}θ}_{a^{2}}u^{\frac{1}{2}}dudθ[/itex]

[itex]=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}(sin^{3}θ-1)dθ[/itex]

[itex]=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}sin^{3}θdθ+\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ[/itex]

sin3θ is an odd function so the first integral is equal to zero:

[itex]=\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ[/itex]

[itex]=\frac{2}{3}πa^{3}[/itex]

However, I have seen other solutions online that give the actual answer as 2a3(3π-4)/9. The authors of these solutions change the limits of integration of the original triple integral by taking advantage of symmetry. More specifically, I noticed that the person changed the limits of θ from 0 to π/2 by multiplying the integral by 2. Then when you get to the point when you integrate sin3θ, the integral no longer equals zero. However, I thought you could only do this if the region of integration has no θ dependence. And in any case, why should it matter whether I change the limits of integration or not? It's still the same integral, right? I have a feeling I'm making a very dumb mistake somewhere but I can't find out where. Thanks for any help.
 
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Answers and Replies

  • #2
DryRun
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Your integral is correct.
sin3θ is an odd function so the first integral is equal to zero:
You could also have used the multiple angle formula to convert sin3θ into sin(3θ) or used the substitution: let u = cosθ.
 
  • #3
LCKurtz
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Hi everyone. I am trying to integrate the following:

[itex]\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ[/itex]

Here's my work:

[itex]=2\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}r\sqrt{a^{2}-r^{2}}drdθ[/itex]

I use substitution with u=a2-r2 to get:

[itex]=-\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{a^{2}sin^{2}θ}_{a^{2}}u^{\frac{1}{2}}dudθ[/itex]

[itex]=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}(sin^{3}θ-1)dθ[/itex]

[itex]=-\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}sin^{3}θdθ+\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ[/itex]

sin3θ is an odd function so the first integral is equal to zero:

[itex]=\frac{2}{3}a^{3}\int^{\frac{π}{2}}_{-\frac{π}{2}}dθ[/itex]

[itex]=\frac{2}{3}πa^{3}[/itex]

However, I have seen other solutions online that give the actual answer as 2a3(3π-4)/9. The authors of these solutions change the limits of integration of the original triple integral by taking advantage of symmetry. More specifically, I noticed that the person changed the limits of θ from 0 to π/2 by multiplying the integral by 2. Then when you get to the point when you integrate sin3θ, the integral no longer equals zero. However, I thought you could only do this if the region of integration has no θ dependence. And in any case, why should it matter whether I change the limits of integration or not? It's still the same integral, right? I have a feeling I'm making a very dumb mistake somewhere but I can't find out where. Thanks for any help.
You didn't make a dumb mistake, but it is a mistake nevertheless, sort of a subtle one. The problem is where you have $$
u^{\frac 3 2}|_{a^2}^{a^2\sin^2\theta}$$and you substitute the ##a^2\sin^2\theta## in for the ##u##. That gives you $$
(a^2\sin^2\theta)^{\frac 3 2}$$That does not simplify to ##a^3\sin^3\theta## when ##\theta## is between ##-\frac \pi 2## and ##0## because ##a^2\sin^2\theta\ge 0## and when you raise it to the 3/2 power it must be nonnegative. It's the old problem that ##\sqrt{x^2}= |x|##, not ##x##. What it does simplify to is ##a^3|\sin^3(\theta)|##, which is not an odd function and doesn't give ##0## in later steps.
 
  • #4
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Thanks LCKurtz. That ,makes sense. Now what I don't understand is this:

[itex]\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ=4\int^{\frac{π}{2}}_{0}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{0}rdzdrdθ[/itex]

It makes sense to me with the integral with respect to z, since the region of integration is symmetrical about the xy-plane, and f(θ,r,z)=r is even with respect to z. As for the integral with respect to θ, I realize that the region of integration is also symmetrical about the yz-plane, but I am not immediately inclined to think that the function f(θ) being integrated at that time will be even with respect to θ. How can you tell? Am I even thinking of symmetry in the right way?

EDIT: Is it because g(θ)=acosθ is even with respect to θ, so any composite function f(g(θ)) that results from the integration with respect to r will also be even?
 
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  • #5
LCKurtz
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Thanks LCKurtz. That ,makes sense. Now what I don't understand is this:

[itex]\int^{\frac{π}{2}}_{-\frac{π}{2}}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{-\sqrt{a^{2}-r^{2}}}rdzdrdθ=4\int^{\frac{π}{2}}_{0}\int^{acosθ}_{0}\int^{\sqrt{a^{2}-r^{2}}}_{0}rdzdrdθ[/itex]

It makes sense to me with the integral with respect to z, since the region of integration is symmetrical about the xy-plane, and f(θ,r,z)=r is even with respect to z. As for the integral with respect to θ, I realize that the region of integration is also symmetrical about the yz-plane, but I am not immediately inclined to think that the function f(θ) being integrated at that time will be even with respect to θ. How can you tell? Am I even thinking of symmetry in the right way?

EDIT: Is it because g(θ)=acosθ is even with respect to θ, so any composite function f(g(θ)) that results from the integration with respect to r will also be even?
##\cos\theta## is even and any function of an even function is even. See if you can prove that.
 
  • #6
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Let x=g(t) be an even function, that is, x=g(t)=g(-t). Then f(x)=f(g(t))=f(g(-t)) must also be an even function.

Is that a sufficient proof? It sounds more like a conclusion.
 
  • #7
LCKurtz
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Let x=g(t) be an even function, that is, x=g(t)=g(-t). Then f(x)=f(g(t))=f(g(-t)) must also be an even function.

Is that a sufficient proof? It sounds more like a conclusion.
That's about all there is to it. I would write it like this: Let ##h(t) = f(g(t))##. You want to show ##h(t) = h(-t)##. So calculate ##h(-t)##:
##h(-t) = f(g(-t)) = f(g(t))\hbox{ (since g is even) } = h(t)##.
 
  • #8
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Okay, thanks again. This problem was driving me insane before but it's all clear now.
 

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