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Double Integrals in polar coordinates setup

  1. Nov 18, 2009 #1
    Use polar coordinates to find the volume of the given solid inside the sphere x^2 +y^2 + z^2 = 16 and outside the cylinder x^2 +y^2 = 4

    I know how to set up the the integral to find the volume inside the sphere but Im not quite sure how to also find the outside of the cylinder. Can someone confirm if this right or wrong?

    x^2 + y^2 +z^2 = 16

    z^2 = 16 - x^2 - y^2

    z = sqrt( 16 - r^2 )

    since the problem asks for volume of the sphere z = 2*sqrt( 16 - r^2 )

    x^2 + y^2 = 4

    r^2 = 4

    r = 2

    so 2 < r < 4, and 0 < theta < 2pi

    are my bounds set up correct?
     
  2. jcsd
  3. Nov 18, 2009 #2

    Dick

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    That seems ok so far.
     
  4. Nov 18, 2009 #3

    Mark44

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    No, since you aren't taking z into account. I'm assuming that you're going to use a triple integral to find this volume. What you would likely get is the volume between a sphere of radius 2 and a sphere of radius 4.

    Because of the symmetry of the cylinder and the sphere, you can take a shortcut and find the volume of the portion of the region in the first octant, and then multiply that result by 8.

    If you haven't drawn a sketch of the region you're integration, you should do so. The circular cylinder is partially inside the sphere.
     
  5. Nov 18, 2009 #4

    Dick

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    I think r_swayze is planning on integrating 2*sqrt(16-r^2) over the domain 2<=r<=4 in 2 dimensional polar coordinates, not spherical coordinates. Admitted, the statement is a little flawed, but that's what I was reading through the lines.
     
  6. Nov 18, 2009 #5

    Mark44

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    I confess to missing the thread title, which clearly states his intention to use double integrals.
     
  7. Nov 18, 2009 #6

    The problem asks to only use double integrals though. I see what you mean by the sphere of radius 2 and 4. How can I account for the cylinder inside the sphere then using double integrals?
     
  8. Nov 18, 2009 #7

    Dick

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    I think you already had it right! Maybe I'll let Mark44 finish this one.
     
  9. Nov 19, 2009 #8

    Mark44

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    r_swayze,
    Yes, you had the description of the region over which integration is taking place correct. To follow up on my earlier comment on the symmetry of both the sphere and the cylinder, I will integrate over the region in the first quadrant between two circles: r = 2 and r = 4, and for [itex]\theta[/itex] ranging between 0 and [itex]\pi/2[/itex]. We need to remember to multiply our result by 8, since the actual solid is above all four quadrants as well as below all four.

    Now to the integration. Imagine two rays extending out from the z-axis in the x-y plane, one at an angle of theta, and the other at an angle of [itex]\theta[/itex] + d[itex]\theta[/itex]. Imagine also that these rays are divided up into subintervals of length dr. We have small area elements that are not quite square, each of area r*dr*[itex]d\theta[/itex]. These elements run along the rays between r = 2 and r = 4. Now, at each of these small area elements, extend a line segment vertically from the x-y plane up to the sphere. Each of these line segments is z = sqrt(16 - r2) in height.

    We now have everything we need for the iterated integral: the integrand and the limits of integration.
    [tex]V~=~8~\int_{\theta = 0}^{\pi/2} \int_{r = 2}^4 \sqrt{16 - r^2}~r~dr~d\theta[/tex]

    I hope that with my explanation you understand where the integrand came from, and that you can finish this off.
     
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