Triple Integral converting from cylindrical to spherical

In summary: Next, use the quadratic formula to solve for \rho. \rho^2=r^2+z^2\left(r^2\right)^2\sin\phi\left(\rho\sin\phi\right) The result is \rho=\frac{z^2}{2r^2}.
  • #1
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Homework Statement



Convert the following integral to an equivalent integral in spherical coordinates.
Do NOT evaluate the integral.

∫∫∫ r^3 dz dr dtheta

limits of integration
pi/4<theta<pi/2
0<r<2
0<z<√(2r-r^2)



Homework Equations



z=pcos(theta)
r^2=x^2 +y^2
p^2=x^2 +y^2 +z^2
theta=theta



The Attempt at a Solution



My problem is not converting the actual integrand but finding the new limits of integration. I realize that the theta limits will stay the same, but I am not sure how to find the rho or phi limits.

I have found through conversions that the new integrand will be

∫∫∫(p^5)sin(phi)^4 dp dphi dtheta
 
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  • #2
That's kina' messy and just cus' he said not doesn't mean you don't have too. First need to get straight the cylindrical and spherical coordinate system. But for starters, just get the answer and then work towards that. You have:

[tex]\int_{\pi/4}^{\pi/2}\int_0^2\int_{0}^{\sqrt{2r-r^2}} f(z,r,\theta) rdzdrd\theta[/tex]

[tex]\int_{\pi/4}^{\pi/2}\int_0^2\int_{0}^{\sqrt{2r-r^2}} \left(r^2\right) rdzdrd\theta\approx 2.15898[/tex]

See, just did it. So anything I come up with in spherical coordinates, has to equal that (numerically in Mathematica or whatever).

Now, if I have a function:

[tex]f(z,r,\theta)=r^2[/tex]

How can I write that in terms of a function in spherical coordinates:

[tex]f(z,r,\theta)=r^2=h(\rho,\theta,\phi)[/tex]

Keep in mind the angle [itex]\phi[/itex] is mearured down from the y-axis where it is zero, to the x-axis where it is [itex]\pi/2[/itex]

So draw some triangles, z, rho, phi and all, and can you compute:

[tex]f(z,r,\theta)=r^2=h(\rho,\theta,\phi)=k(\rho,\phi)[/tex]

I'll get you started. Surely, we can write between the two coordinate systems along a typical triangle in both spaces:

[tex]\rho^2=r^2+z^2[/tex]

for the angle [itex]\pi/2-\phi[/itex]

so from just that relation, can you come up with an expression for [itex]r^2[/itex] in terms of [itex]\rho[/itex] and [itex]\phi[/itex]?
 
  • #3
After converting r^3 from cylindrical to spherical I attained the value of (p^5)*sin(phi)^4, I still don't understand how to get the new limits of integration though
 
  • #4
Normally you would use a combination of algebra and a picture to figure out the limits. I would suggest that you first plot [itex]z =\sqrt{2r-r^2}[/itex] in a z-r plane for r from 0 to 2. Your cylindrical equation doesn't depend on [itex]\theta[/itex] so the surface is that graph rotated through [itex]\frac \pi 4[/itex]. That should help you figure out the [itex]\phi[/itex] limits.

To figure out the [itex]\rho[/itex] limits you need to express your equation in spherical coordinates and solve for [itex]\rho[/itex]. You know the spherical coordinates for z and remember [itex]r =\rho\sin\phi[/itex]. Start by squaring your equation and substituting.
 

What is a triple integral?

A triple integral is a type of mathematical operation used in multivariable calculus to calculate the volume of a three-dimensional region. It involves integrating a function over a three-dimensional space.

What is the difference between cylindrical and spherical coordinates?

Cylindrical coordinates are a type of coordinate system that uses a distance from the origin, an angle in the xy-plane, and a height above the xy-plane to specify a point in three-dimensional space. Spherical coordinates use a distance from the origin, an angle in the xy-plane, and an angle from the positive z-axis to specify a point in three-dimensional space.

Why would I need to convert from cylindrical to spherical coordinates for a triple integral?

In some situations, the shape of the region being integrated over may be better described using spherical coordinates rather than cylindrical coordinates. Converting between the two coordinate systems allows for easier calculation of the triple integral.

How do I convert a triple integral from cylindrical to spherical coordinates?

To convert a triple integral from cylindrical to spherical coordinates, you need to substitute the expressions for the cylindrical coordinates (r, θ, z) with the corresponding spherical coordinates (ρ, φ, θ). This will change the limits of integration and the integrand.

Are there any common mistakes to avoid when converting a triple integral from cylindrical to spherical coordinates?

One common mistake is forgetting to change the limits of integration when converting between coordinate systems. It is also important to make sure the integrand is in the correct form for the new coordinate system. Additionally, it is important to pay attention to the orientation of the coordinate axes and their corresponding angles when making the conversion.

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