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Thread: Not Able To Obtain The Same Answer In Spherical/Cylindrical Coordinate

  1. #1
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    Not Able To Obtain The Same Answer In Spherical/Cylindrical Coordinate

    Hi,

    I have to solve this problem using spherical coordinate and validate my answer using cylindrical coordinate, but I have 2 different answers, obviously!

    Can anyone tell me what I am doing wrong? (See attached file for the problem and what I did)

    Thanks, Mike.

    EDIT: sorry, I used the wrong angle for the limit of $\displaystyle \varphi$, it should be $\displaystyle \frac{\pi}{6}$ not $\displaystyle \frac{\pi}{3}$.

    My new answer is 0,420894

    EDIT 2: I drew the solid in my cad software and came up with a volume of 0,561.

    I tried to do it the long way, calculating the volume of the cone, sphere and cylinder.

    H of cone (z) = $\displaystyle 4-\sqrt{3(0^2+2^2)}=4-.535898=3.4641$

    Cone: V=$\displaystyle (\frac{1}{3}\pi*2^2*3.4641)=14.5104$
    Sphere: V=$\displaystyle (\frac{4}{3}\pi*2^3)=\frac{16\pi}{3}$
    Cylinder: V=$\displaystyle (\pi*2^2*.535898)=6.7343$

    Total volume = Cone + Cylinder - Sphere = 14.5104 -$\displaystyle \frac{16\pi}{3}$ + 6.7343 = 4,48953
    Since I only need from $\displaystyle \frac{\pi}{4}$ to $\displaystyle \frac{\pi}{2}$ (one eight) and not 2$\displaystyle \pi$, I divide by 8.

    $\displaystyle \frac{4,48953}{8}=0,561191$
    Attached Files Attached Files
    Last edited by MoneyHypeMike; Jul 3rd 2011 at 07:57 PM. Reason: Adding information
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  2. #2
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    Re: Not Able To Obtain The Same Answer In Spherical/Cylindrical Coordinate

    For the spherical coordinates case, your limits for $\displaystyle \rho$ look wrong. Firstly, if you're above the sphere, the lower limit should be 2. Secondly, you have not converted the Cartesian equation for the cone correctly into sphericals. The equation $\displaystyle \phi =\pi/3$ is a cone with vertex at the origin; this is not what you were given.
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