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Changing to spherical coordinates

  • Thread starter planck42
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  • #1
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Homework Statement


Evaluate by changing to spherical coordinates
[tex]\int^1_0\int^{\sqrt{1-x^2}}_0\int^{\sqrt{2-x^{2}-y^2}}_{\sqrt{x^{2}+y^2}}xydzdydx[/tex]


Homework Equations


[tex]dz dy dz = {\rho}^{2}sin{\phi}[/tex]


The Attempt at a Solution


This problem is quite simple to do in cylindrical coordinates, which I believe gives 0 because of a [tex]sin{\theta}cos{\theta}[/tex] term being integrated from 0 to 2pi, but the request is to use spherical coordinates. The bounds on the integral make a cylinder with radius 1 sandwiched between two cones. Since those bounds obviously don't have spherical symmetry, I am lost as to how to make the conversion.
 

Answers and Replies

  • #2
tiny-tim
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hi planck42! :wink:

show us how how far you can get, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
SammyS
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Homework Statement


Evaluate by changing to spherical coordinates
[tex]\int^1_0\int^{\sqrt{1-x^2}}_0\int^{\sqrt{2-x^{2}-y^2}}_{\sqrt{x^{2}+y^2}}xydzdydx[/tex]


Homework Equations


[tex]dz dy dz = {\rho}^{2}sin{\phi}[/tex]


The Attempt at a Solution


This problem is quite simple to do in cylindrical coordinates, which I believe gives 0 because of a [tex]sin{\theta}cos{\theta}[/tex] term being integrated from 0 to 2pi, but the request is to use spherical coordinates. The bounds on the integral make a cylinder with radius 1 sandwiched between two cones. Since those bounds obviously don't have spherical symmetry, I am lost as to how to make the conversion.
[tex]dz\,dy\,dz={\rho}^{2}\sin{\phi}\,d\rho\,d\phi\,d\theta[/tex]

For the integration over z, the lower integration limit is:

[tex]z=\sqrt{x^{2}+y^2}\,,[/tex] which is a cone.

The upper limit is:

[tex]\textstyle z=\sqrt{2-x^{2}-y^2}\ \ \to\ \ x^{2}+y^2+z^2=2\,,[/tex] which is the upper half of a sphere centered at the origin having radius [tex]\textstyle \sqrt{2}\,.[/tex]

These surfaces meet at the circle, [tex]\textstyle x^{2}+y^2=1,\ z=1\,.[/tex]

Spherical coordinates are a natural for this integral.

The limits of integration for y, indicate that 0 ≤ θπ.
 
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  • #4
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hi planck42! :wink:

show us how how far you can get, and where you're stuck, and then we'll know how to help! :smile:
First off, the restrictions on x and y make it clear that theta goes from 0 to pi/2.
Next, phi goes from 0 on the sphere which was pointed out by SammyS(I didn't notice that initially, but it's a tremendous boost) to pi/4 on the cone.
That leaves rho, which ends at sqrt 2 and begins... where exactly?
 
  • #5
tiny-tim
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hi planck42! :smile:

to find the limits for ρ,

choose fixed values of θ and φ, and find the limits of ρ for those fixed values :wink:
 
  • #6
SammyS
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First off, the restrictions on x and y make it clear that theta goes from 0 to pi/2.
Next, phi goes from 0 on the sphere which was pointed out by SammyS(I didn't notice that initially, but it's a tremendous boost) to pi/4 on the cone.
That leaves rho, which ends at sqrt 2 and begins... where exactly?
Yes, I agree. When I stated that "The limits of integration for y, indicate that 0 ≤ θπ.", I only meant to indicate that θ lies between 0 and π, not that those are the limits of integration for θ. As you (planck42) point out, the added restrictions imposed by x, further restrict θ.

The region over which you are integrating is a section of a http://mathworld.wolfram.com/SphericalCone.html" [Broken])

As for ρ, look at the lower limits for x and y. Plug those into the lower limit for z.
 
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  • #7
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I just got a better idea: when the cone hits the cylinder, z=r=1, but conveniently z=1 is where the cone intersects the sphere! Therefore, a dome-like shape is the region of integration, and if one were to move the xy-plane up to the circle which is the base of the dome, then it would get rid of my issue with identifying the lower bound for rho in exchange for a bit of complications on the upper bound. http://mathworld.wolfram.com/Ellipsoid.html has everything on this particular shape, with a^2=b^2=1 in this case.

Answer pending
 
  • #8
SammyS
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I just got a better idea: when the cone hits the cylinder, z=r=1, but conveniently z=1 is where the cone intersects the sphere! Therefore, a dome-like shape is the region of integration, and if one were to move the xy-plane up to the circle which is the base of the dome, then it would get rid of my issue with identifying the lower bound for rho in exchange for a bit of complications on the upper bound. http://mathworld.wolfram.com/Ellipsoid.html has everything on this particular shape, with a^2=b^2=1 in this case.

Answer pending
The vertex of the cone is at (x,y,z)=(0,0,0).

∴ , the vertex of the cone is at ρ=0.
 
  • #9
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The vertex of the cone is at (x,y,z)=(0,0,0).

∴ , the vertex of the cone is at ρ=0.
What does that have to do with anything? If you're implying that rho be from 0 to 2, then that's wrong because rho from 0 to two means the entire region under the sphere of radius 2. By the way, the answer I got was this

[tex]\frac{1}{10}\int^\frac{\pi}{2}_{0}\sin^{3}{\phi}(\sin^{2}{\phi}+\frac{{\cos}^{2}{\phi}}{c^{2}})^{-\frac{5}{2}}d{\phi}[/tex]

where [tex]c^{2}=3-2\sqrt{2}[/tex]

I used the equation [tex]x^{2}+y^{2}+\frac{z^2}{c^2}=1, c=\sqrt{2}-1[/tex], which translates to [tex]{\rho}=(\sin^{2}{\phi}+\frac{\cos^{2}{\phi}}{c^2}[/tex] and worked out the rho and theta integrals. Needless to say, it requires a calculator, which spews out roughly .276 as the value of the integral. I suppose that makes sense, given that there's not much region between the sphere and the cone.
 
  • #10
SammyS
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What does that have to do with anything? If you're implying that rho be from 0 to 2, then that's wrong because rho from 0 to two means the entire region under the sphere of radius 2. By the way, the answer I got was this

[tex]\frac{1}{10}\int^\frac{\pi}{2}_{0}\sin^{3}{\phi}(\sin^{2}{\phi}+\frac{{\cos}^{2}{\phi}}{c^{2}})^{-\frac{5}{2}}d{\phi}[/tex]

where [tex]c^{2}=3-2\sqrt{2}[/tex]

I used the equation [tex]x^{2}+y^{2}+\frac{z^2}{c^2}=1, c=\sqrt{2}-1[/tex], which translates to [tex]{\rho}=(\sin^{2}{\phi}+\frac{\cos^{2}{\phi}}{c^2}[/tex] and worked out the rho and theta integrals. Needless to say, it requires a calculator, which spews out roughly .276 as the value of the integral. I suppose that makes sense, given that there's not much region between the sphere and the cone.
I was implying that the lower limit for the ρ integration is zero.

The upper limit is set by the spherical surface:

[tex]
\textstyle x^{2}+y^2+z^2=2\,\ \ \to\ \ \rho^2=2\,.
[/tex] ← which I thought you previously determined.
 
  • #11
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I was implying that the lower limit for the ρ integration is zero.

The upper limit is set by the spherical surface:

[tex]
\textstyle x^{2}+y^2+z^2=2\,\ \ \to\ \ \rho^2=2\,.
[/tex] ← which I thought you previously determined.
I don't quite understand what you're trying to say; are you referring to the bounds in the translated or to those in the untranslated coordinate system? If the former, then I agree [that the lower bound is 0] but the upper bound is no longer sqrt 2, and it can't be the latter.
 
  • #12
SammyS
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This is what I get:

[tex]
\int_0^1\int_0^{\sqrt{1-x^2}}\int_{\sqrt{x^{2}+y^2}}^{\sqrt{2-x^{2}-y^2}}\,xy\ dz\,dy\,dx
[/tex]


[tex]
=\int_0^{\pi/4}\int_0^{\pi/2}\int_0^{\sqrt{2}} {\rho}^{4}\sin^3{\phi}\,\sin\theta\,\cos\theta\,d\rho\,d\theta\,d\phi
[/tex]​
 
  • #13
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This is what I get:

[tex]
\int_0^1\int_0^{\sqrt{1-x^2}}\int_{\sqrt{x^{2}+y^2}}^{\sqrt{2-x^{2}-y^2}}\,xy\ dz\,dy\,dx
[/tex]


[tex]
=\int_0^{\pi/4}\int_0^{\pi/2}\int_0^{\sqrt{2}} {\rho}^{4}\sin^3{\phi}\,\sin\theta\,\cos\theta\,d\rho\,d\theta\,d\phi
[/tex]​
Why is the lower bound for rho zero? The sphere meets the cone well before that.
 
  • #14
SammyS
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The vertex of the cone: z = √(x2 + y2) is at the origin.

The sphere which the cone meets is given by ρ = √(2) in spherical coordinates, which is one reason why the integral works out so nicely in spherical coordinates.

The cone is given by ɸ = π/4 .
 
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  • #15
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The vertex of the cone: z = √(x2 + y2) is at the origin.

The sphere which the cone meets is given by ρ = √(2) in spherical coordinates, which is one reason why the integral works out so nicely in spherical coordinates.

The cone is given by ɸ = π/4 .
Oh duh the cone isn't a bound on rho. In that case, I agree with the integral you've got and arrived at an answer of [tex]\frac{1}{60}(8\sqrt{2}-10)[/tex], which is approximately 0.022
 

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