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Math Help - Triple integral

  1. #1
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    Triple integral

    Please, I need some help to evaluate integral:
    \int \frac {z^2}{(x^2+y^2+z^2)^\frac{3}{2}} dxdydz
    in area that is bounded with
    4z \geq x^2+y^2+z^2+(8/3)
    z\geq\sqrt{3x^2+3y^2}
    Thanks in advance.
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  2. #2
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    Quote Originally Posted by TCKony View Post
    Please, I need some help to evaluate integral:
    \int \frac {z^2}{(x^2+y^2+z^2)^\frac{3}{2}} dxdydz
    in area that is bounded with
    4z \geq x^2+y^2+z^2+(8/3)
    z\geq\sqrt{3x^2+3y^2}
    Thanks in advance.
    Have you drawn the region?
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  3. #3
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    4z= x^2+ y^2+ z^2+ (8/3) is the same as x^2+ y^2+ z^2- 4z+ 4- 4+ 8/3= x^2+ y^2+ (z- 2)^2- (4/3)= 0, a sphere with center at (0, 0, 2) and radius 2/\sqrt{3}. 4zk\ge x^2+ y^2+ z^2+ (8/3) is the same as x^2+ y^2+ z^2- 4z+ 4- 4+ 8/3= x^2+ y^2+ (z- 2)^2\le (4/3), the region on and inside that sphere. z= \sqrt{3x^2+ 3y^2}= \sqrt{3}\sqrt{3} is the same as z= \sqrt{3}r where r is the radius is the xy-plane in any direction. In other words, it is the upper part of a cone- the line z= \sqrt{3}x rotated around the z- axis. z\ge \sqrt{3x^2+3y^2} is the region above (interior to) that cone.
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  4. #4
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    I have drawn that, but I need the result to check my solution, that is \frac {\pi*373}{270\sqrt{3}}.
    Thanks.
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