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Math Help - Triple Integral, Spherical Coordinates

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    Triple Integral, Spherical Coordinates

    I am confused on how the bounds on the question are found. Here's the question.

    Use spherical coordinates to find the mass of the solid enclosed by the hemispherical region x^2+y^2+z^2<4 and 0<y with a density function (x,y,z)=
     \sqrt{1+(x^2+y^2+z^2)^\frac{3}{2})}

    also, the < signs are suppose to be "less than or equal to"
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    Hello,

    The bound for y will be 0 and 2 :
    0<y<4-x-z<4

    y>0
    -> y<sqrt(4)=2

    0<y<2

    Are there any information about x or z ?
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    Nope, thats all the info given.
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    Quote Originally Posted by Gotovina7 View Post
    I am confused on how the bounds on the question are found. Here's the question.

    Use spherical coordinates to find the mass of the solid enclosed by the hemispherical region x^2+y^2+z^2<4 and 0<y with a density function (x,y,z)=
     \sqrt{1+(x^2+y^2+z^2)^\frac{3}{2})}

    also, the < signs are suppose to be "less than or equal to"
    Looks to me like the hemisphere is the right half of a sphere centred on the origin.

    You're told to work in spherical coordinates, right? So:

    Radius: 0 \leq r \leq 4.

    Azimuthal angle in the xy-plane: -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.

    Polar angle from the z-axis: 0 \leq \phi \leq \pi.

    Of interest: Spherical Coordinates
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    Since {\rho}^{2}=x^{2}+y^{2}+z^{2}

    We can sub it into the density function and get

    \sqrt{1+{\rho}^{3}}
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    I am just a bit confused on how you find the "phi" and "theta" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful
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    I think your set up should look something like this:

    \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}
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    Quote Originally Posted by galactus View Post
    I think your set up should look something like this:

    \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}
    Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it.

    EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and  0<y is. What do these regions mean?
    Last edited by Gotovina7; April 5th 2008 at 05:35 PM.
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    Quote Originally Posted by galactus View Post
    I think your set up should look something like this:

    \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}
    Ahh yes. I misread my x- and y-axes. 0 \leq \theta \pi indeed.
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  10. #10
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    Quote Originally Posted by Gotovina7 View Post
    Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it.

    EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and  0<y is. What do these regions mean?
    You're told it's a hemisphere in the question: " ....enclosed by the hemispherical region ...."

    Furthermore:

    Quote Originally Posted by mr fantastic View Post
    Looks to me like the hemisphere is the right half of a sphere centred on the origin.

    [snip]
    because 0 < y => y > 0.
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