Setting up a triple integral using cylindrical & spherical coordinates

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SUMMARY

This discussion focuses on setting up a triple integral for a region defined by the sphere \(x^2 + y^2 + z^2 = R^2\) and the planes \(z = \frac{R}{2}\) and \(z = R\) using cylindrical and spherical coordinates. The correct upper limit for \(r\) in cylindrical coordinates is confirmed as \(R\sqrt{\frac{3}{4}}\), while the lower limit for \(z\) is established as \(z = \frac{R}{2}\). In spherical coordinates, the integral setup involves \(\rho\) depending on \(\phi\), and the integral is expressed as \(\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_{\frac{R}{2}}^R \rho^{2}\sin\theta\,d\rho\,d\phi\,d\theta\).

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Knowledge of cylindrical coordinates and their application
  • Familiarity with spherical coordinates and transformations
  • Ability to interpret equations of planes and spheres in three-dimensional space
NEXT STEPS
  • Study the derivation of limits for triple integrals in cylindrical coordinates
  • Learn about spherical coordinate transformations and their applications in integration
  • Explore the concept of volume elements in different coordinate systems
  • Practice solving triple integrals with varying limits based on geometric constraints
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calcuseless
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Homework Statement



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

Homework Equations



\iiint\limits_Gr\,dz\,dr\,d\theta

\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta

The Attempt at a Solution



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

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

Homework Statement



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

Homework Equations



\iiint\limits_Gr\,dz\,dr\,d\theta

\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta

The Attempt at a Solution



2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta
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.

\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta
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 \rho. You will find that \rho depends on \phi, but that's OK.
 
LCKurtz said:
You will find that \rho depends on \phi, but that's OK.

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

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
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.
 
calcuseless said:
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
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:
 

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