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Math Help - Spherical Coordinate problem

  1. #1
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    Spherical Coordinate problem

    the problem reads "Find the volume of the part of the ball p(rho)=<a that lies between the cones phi=pi/6 and phi=pi/3"

    can anyone help at least set up the triple integral? I'm not sure how to go about this
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by RoadRaider12 View Post
    the problem reads "Find the volume of the part of the ball p(rho)=<a that lies between the cones phi=pi/6 and phi=pi/3"

    can anyone help at least set up the triple integral? I'm not sure how to go about this
    From the question I gather the equation for the sphere(ball) is

    x^2 + y^2 + z^2 \le ( \sqrt{a} ) ^2

    What we're looking for is

    \iiint dV where dV = p^2 sin \phi dp d \phi d \theta

    So what are our bounds for theta, phi and p? Well, for P this is the value of the minimum point and maximum point. In this case, clearly the farthest point P is  \sqrt{a} . But what is the shortest point/distance it can be? Well, in this case the origin! So,

    0 \le P \le \sqrt{a}

    From the question were are given our bounds of phi!

     \frac{ \pi }{6} \le \phi \le \frac{ \pi }{3}

    What about theta? Well, we are going 360 degress around the way! So our theta bounds become

     0 \le \pi \le 2 \pi

    Therefore,

    \int_0^{2 \pi } d \theta \int_{ \frac{ \pi }{6} }^{ \frac{ \pi }{3} } sin \phi d \phi \int_0^{ \sqrt{a} } p^2 dp
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