Setting up a triple integral using cylindrical & spherical coordinates

In summary, the problem asks to show the region inside a sphere and between two planes in cylindrical and spherical coordinates. The equations for the volume elements in cylindrical and spherical coordinates are given, and the problem involves finding the lower limits for z and \rho in each coordinate system. The lower limit for z can be found by expressing the equation of the plane z = R/2 in spherical coordinates and solving for \rho, while the lower limit for \rho can be found by solving the same equation for \rho and recognizing that it depends on \phi. The 2 in front of the integrals is not necessary.
  • #1
calcuseless
9
0

Homework Statement



Inside the sphere x2 + y2 + z2 = R2 and between the planes z = [tex]\frac{R}{2}[/tex] and z = R. Show in cylindrical and spherical coordinates.

Homework Equations



[tex]\iiint\limits_Gr\,dz\,dr\,d\theta[/tex]

[tex]\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]

The Attempt at a Solution



[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is my upper limit for r correct? How do I find the lower limit for z?

[tex]\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]
How do I find the lower limit for rho?
 
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  • #2
calcuseless said:

Homework Statement



Inside the sphere x2 + y2 + z2 = R2 and between the planes z = [tex]\frac{R}{2}[/tex] and z = R. Show in cylindrical and spherical coordinates.

Homework Equations



[tex]\iiint\limits_Gr\,dz\,dr\,d\theta[/tex]

[tex]\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]

The Attempt at a Solution



[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is my upper limit for r correct? How do I find the lower limit for z?

Yes the upper limit is OK. Isn't the equation of the plane z = R/2? That would be z on the lower surface.

[tex]\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]
How do I find the lower limit for rho?

Again, the equation of that plane is z = R/2. Express that in spherical coordinates and solve it for [itex]\rho[/itex]. You will find that [itex]\rho[/itex] depends on [itex]\phi[/itex], but that's OK.
 
  • #3
LCKurtz said:
You will find that [itex]\rho[/itex] depends on [itex]\phi[/itex], but that's OK.

Ah, I didn't understand that concept. Thank you!

[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.
 
  • #4
calcuseless said:
[tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.

I don't know why you put it there either. Go with your instincts. If you don't think it should be there, don't put it there. :cool:
 

Related to Setting up a triple integral using cylindrical & spherical coordinates

1. What are the steps to set up a triple integral using cylindrical coordinates?

To set up a triple integral using cylindrical coordinates, you first need to identify the bounds of the integral. These will be dependent on the shape of the region you are integrating over. Next, you need to convert the bounds from Cartesian coordinates to cylindrical coordinates. Finally, you can write the integral in the form of ∫∫∫f(r,θ,z)rdrdθdz, where f(r,θ,z) is the function being integrated.

2. How do I convert between Cartesian and cylindrical coordinates?

To convert between Cartesian and cylindrical coordinates, you can use the following equations: x = rcosθ, y = rsinθ, and z = z. The inverse equations are r = √(x²+y²), θ = arctan(y/x), and z = z. It is important to note that r is the distance from the origin to the point in the xy-plane, and θ is the angle between the positive x-axis and the line connecting the origin to the point.

3. What are the advantages of using cylindrical coordinates in triple integrals?

One advantage of using cylindrical coordinates in triple integrals is that they are well-suited for integrating over regions with circular or cylindrical symmetry. This can make the integration process much simpler and more efficient. Additionally, cylindrical coordinates can often lead to simpler and more elegant solutions compared to using Cartesian coordinates.

4. When should I use spherical coordinates instead of cylindrical coordinates?

Spherical coordinates are typically used when the region being integrated over has spherical symmetry. This means that the region is symmetric about a point and can be described using a radius and two angles. Spherical coordinates can also be useful when the function being integrated involves expressions with inverse trigonometric functions, as these can be simplified in spherical coordinates.

5. How do I interpret the bounds in a triple integral using spherical coordinates?

In a triple integral using spherical coordinates, the bounds for r, θ, and φ will depend on the shape of the region being integrated over. The bounds for r will typically be from 0 to a certain radius, the bounds for θ will be from 0 to 2π, and the bounds for φ will be from 0 to π. It is important to visualize the region and understand how the bounds correspond to the shape in order to correctly set up the integral.

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