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Thread: Double integral to find the volume of a solid bound by a sphere and cone

  1. #1
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    Double integral to find the volume of a solid bound by a sphere and cone

    Find the volume of the solid bounded by the upper hemisphere
    x^2+y^2+z^2=8 z>0

    and the cone z=squareroot(x^2 + y^2).
    What is the double double integral i need to find the volume, what would the limits be? It didnt seem that hard until i actually started doing it. thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Without using a double integral, it is possible to find the volume by splitting it into two volumes which can be easily computed :

    The intersection of the sphere and of the cone is given by $\displaystyle \left\{ \begin{array}{l} x^2+y^2+z^2=8 \\ z=\sqrt{x^2+y^2}\end{array}\right.\implies z=2$

    If $\displaystyle z\in[0,\,2]$, $\displaystyle z=\sqrt{x^2+y^2}\in [0,\,2]$ hence $\displaystyle x^2+y^2+z^2\leq 8$ : the part of the cone such that $\displaystyle z\in[0,\,2]$ is inside the sphere.

    If $\displaystyle z>2$, $\displaystyle x^2+y^2=8-z^2<4$ hence $\displaystyle \sqrt{x^2+y^2}<2$ : the part of the sphere such that $\displaystyle z>2$ is inside the cone.

    Thus the volume we're looking for is the sum of the volume bounded by the cone for $\displaystyle z\in[0,\,2]$ (red surface, see attached images) and of the volume bounded by the sphere for $\displaystyle z>2$. (green surface) Can you take it from here ?
    Attached Thumbnails Attached Thumbnails Double integral to find the volume of a solid bound by a sphere and cone-cone_sphere_1.jpg   Double integral to find the volume of a solid bound by a sphere and cone-cone_sphere_2.jpg  
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  3. #3
    Eater of Worlds
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    I can make things worse and show you a triple integral in spherical coordinates.

    In the spherical, the equation of the sphere is $\displaystyle {\rho}=2\sqrt{2}$

    And the cone is $\displaystyle {\rho}cos{\phi}=\sqrt{{\rho}^{2}sin^{2}{\phi}cos^{ 2}{\theta}+{\rho}^{2}sin^{2}{\phi}sin^{2}{\theta}}$

    I know this looks menacing, but it whittles down to

    $\displaystyle {\rho}cos{\phi}={\rho}sin{\phi}$

    If we divide both sides by $\displaystyle {\rho}cos{\phi}$, then we get

    $\displaystyle tan{\phi}=1, \;\ {\phi}=\frac{\pi}{4}$

    So, we have the triple integral:

    $\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{2 \sqrt{2}}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d{\theta }$

    It's not that bad to do. Don't ya' wish they covered quadruple integrals in clac class
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