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Evaluating a triple integral in spherical coordinates watch

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    Q: The points of intersection of the cone  z^2 = x^2 + y^2 and the sphere x^2 + y^2 + z^2 = 4z form a circle (in addition to the origin). Find the volume of the solid that lies above the plane containing this circle and inside the sphere.

    I've been trying to solve this question, but can't get the correct answer. The variables in spherical coordinates are \rho,\ \phi, and \theta, where x = \rho \sin \phi \cos \theta, y = \rho \sin \phi \sin \theta and z = \rho \cos \phi. I don't have any tool that could help me draw the surfaces, so you'd have to imagine the surfaces yourself. Solving x^2 + y^2 + z^2 = 4z in spherical coordinates gives me the limits of \rho, i.e from 0 to 4\cos \phi. Solving  z^2 = x^2 + y^2 gives the limits of \phi, i.e from 0 to pi/4. And theta would, of course, be from 0 to 2pi as the region covered is a complete circle.

    Using these limits, the expression for volume should be \displaystyle \int_0^{2\pi} \displaystyle \int_0^{\frac{\pi}{4}} \displaystyle \int_0^{4\cos \phi} \rho^2 \sin \phi\ d\rho\ d\phi\ d\theta.

    Thanks a lot for giving your time to reading the post. I just want to ask where I'm going wrong in getting the correct expression for the volume. ANY help would be appreciated.
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    You haven't considered the plane in which the circle of intersection lies yet (which consequently means that your lower limit for rho is not quite right).

    Also, I don't know follow what you mean by "solving …" before you find the limits - for a volume, you should be looking to solve inequalities, not equations. Visualising the volume will help you to see why your lower rho limit is wrong -- it is the solid section of the sphere that lies above the plane of intersection.
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    (Original post by Zishi)
    Q: The points of intersection of the cone  z^2 = x^2 + y^2 and the sphere x^2 + y^2 + z^2 = 4z form a circle (in addition to the origin). Find the volume of the solid that lies above the plane containing this circle and inside the sphere.

    I've been trying to solve this question, but can't get the correct answer. The variables in spherical coordinates are \rho,\ \phi, and \theta, where x = \rho \sin \phi \cos \theta, y = \rho \sin \phi \sin \theta and z = \rho \cos \phi. I don't have any tool that could help me draw the surfaces, so you'd have to imagine the surfaces yourself. Solving x^2 + y^2 + z^2 = 4z in spherical coordinates gives me the limits of \rho, i.e from 0 to 4\cos \phi. Solving  z^2 = x^2 + y^2 gives the limits of \phi, i.e from 0 to pi/4. And theta would, of course, be from 0 to 2pi as the region covered is a complete circle.

    Using these limits, the expression for volume should be \displaystyle \int_0^{2\pi} \displaystyle \int_0^{\frac{\pi}{4}} \displaystyle \int_0^{4\cos \phi} \rho^2 \sin \phi\ d\rho\ d\phi\ d\theta.

    Thanks a lot for giving your time to reading the post. I just want to ask where I'm going wrong in getting the correct expression for the volume. ANY help would be appreciated.
    For intersection solving the system you will get the z=2 plane
    As the midpont of the spere is (0,0,2) consider u=z-2 transform
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    (Original post by Farhan.Hanif93)
    You haven't considered the plane in which the circle of intersection lies yet (which consequently means that your lower limit for rho is not quite right).

    Also, I don't know follow what you mean by "solving …" before you find the limits - for a volume, you should be looking to solve inequalities, not equations. Visualising the volume will help you to see why your lower rho limit is wrong -- it is the solid section of the sphere that lies above the plane of intersection.
    Alright. By solving, I mean changing the equations into spherical coordinates, i.e replacing x, y and z by \rho, \phi and \theta. Putting x^2 + y^2 = z^2 into x^2 + y^2 + z^2 = 4z gives z = 2, i.e \rho = 2\sec \phi. So my lower limit should be \rho = 2\sec \phi, right?


    (Original post by ztibor)
    For intersection solving the system you will get the z=2 plane
    As the midpont of the spere is (0,0,2) consider u=z-2 transform
    z=2 gives \rho = 2\sec \phi, which should be my lower limit, right?
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    (Original post by Zishi)
    Alright. By solving, I mean changing the equations into spherical coordinates, i.e replacing x, y and z by \rho, \phi and \theta. Putting x^2 + y^2 = z^2 into x^2 + y^2 + z^2 = 4z gives z = 2, i.e \rho = 2\sec \phi. So my lower limit should be \rho = 2\sec \phi, right?



    z=2 gives \rho = 2\sec \phi, which should be my lower limit, right?
    Yes. And the upper?
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    (Original post by ztibor)
    Yes. And the upper?
    \rho = 4\cos \phi?
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    (Original post by ztibor)
    Yes. And the upper?
    BUMP! You haven't replied to post #6 yet. :rolleyes:
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    you learning calculus 3 currently in the u.s?
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    (Original post by Zishi)
    BUMP! You haven't replied to post #6 yet. :rolleyes:
    Sorry
    Your answer is correct.
 
 
 
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