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Math Help - 3 Dimensional Volume Integral

  1. #1
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    3 Dimensional Volume Integral

    Find:

    triple integral ( ?-?; ?-?;?-R) x^2 + y^2 + z^2 dV

    where R is the region that lies above the cone z =c sqrt (x^2 + y^2) and inside the sphere x^2 + y^2 + z^2 = a^2


    no clue really except it needs to be set up in polar form
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  2. #2
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    Quote Originally Posted by hammer View Post
    Find:

    triple integral ( ?-?; ?-?;?-R) x^2 + y^2 + z^2 dV

    where R is the region that lies above the cone z =c sqrt (x^2 + y^2) and inside the sphere x^2 + y^2 + z^2 = a^2


    no clue really except it needs to be set up in polar form
    Not polar form. It needs to be spherical form.

    Let us assume a>0 and c>0.

    Then the we need to consider the following:
    1)How much does theta move?
    2)What is the radius?
    3)What much does phi move?

    Answers:
    1)Theta is from 0 to 2pi because it is a complete revolution.
    2)The radius is a, because x^2+y^2+z^2=a.
    3)We have a cone, the equation of the cone is simply, phi=constant.
    But we need to find what that constant is.
    We know the equation of cone in rectangular coordinates is,
    z=c*sqrt(x^2+y^2)
    z=c*sqrt(r^2)
    z=c*r
    Convert to Spherical,
    rho*sin(phi) = c*rho*cos(phi)
    sin(phi) = c*cos(phi)
    tan(phi)=c
    phi=arctan(c).

    Look at picture below.

    I will do the integral in the next post.
    Attached Thumbnails Attached Thumbnails 3 Dimensional Volume Integral-picture7.gif  
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  3. #3
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    Hier.
    Attached Thumbnails Attached Thumbnails 3 Dimensional Volume Integral-picture8.gif  
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