Use polar coordinates to find the volume of the given solid.

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SUMMARY

The discussion focuses on calculating the volume of a solid defined by the sphere \(x^2 + y^2 + z^2 = 16\) and the cylinder \(x^2 + y^2 = 4\) using polar coordinates. The user initially set the bounds for \(r\) from 2 to 4 and \(\theta\) from 0 to \(2\pi\), leading to a volume calculation of \(16\sqrt{3}\pi\). However, the correct volume is twice this value due to the symmetry of the solid, necessitating an additional integral for the lower half of the sphere. The user clarified that the evaluation of \(dz\) should occur first to account for both halves of the sphere.

PREREQUISITES
  • Understanding of polar coordinates in three dimensions
  • Familiarity with double and triple integrals
  • Knowledge of spherical and cylindrical coordinate systems
  • Basic concepts of volume calculation in calculus
NEXT STEPS
  • Study the application of triple integrals in spherical coordinates
  • Learn about symmetry in volume calculations
  • Explore the use of polar coordinates for area and volume integration
  • Review examples of volume calculations involving both spheres and cylinders
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Students studying calculus, particularly those focusing on multivariable integration, as well as educators looking for examples of volume calculations involving polar coordinates.

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Homework Statement



1. Use polar coordinates to find the volume of the given solid.
2. Inside the sphere x^2 + y^2 + z^2 = 16 and outside the cylinder x^2 + y^2 = 4.

2. The attempt at a solution

My attempt as following:
2<=r<=4, and 0<=theta<=2pi

So I do a double integral of f(x,y)=sqrt(16-r^2)r dr d(theta) and it gives me 16sqrt(3)pi

but I saw the answer somewhere, the volume is 2 times the answer I got, I cannot figure out why 2 times? and in the answer it says "by symmetry", any explanation will be appreciated.
 
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you found the area of the annulus in the polar plane (two integrals: rdrdt), but you need to set your upper and lower bounds wrt the upper and lower halves of the sphere(third integral dz). the double negatives sum to twice your value.

[edit] a little backwards, dz is being evaluated first.
 
Last edited:
Thanks, I think I got the idea here, when I set z=0 I get the sphere on xy plane together with the cylinder to find the range of r,

But I just forget the lower half, that's where the 2 times comes from . :(
 

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