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Math Help - triple integral in given region

  1. #1
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    triple integral in given region

    Evaluate the triple integral where is the solid bounded by the paraboloid and .
    what are the limit of z,y and x to integrate?

    I got
    -sqrt(1-y^2) \leq z \leq sqrt(1-y^2)
    -1 \leq y \leq 1
    0 \leq x \leq 9

    help me
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  2. #2
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    hi


    I perfer to transform the rectangular coordinates to polar ( cylindrial )

    but before changing  dxdydz into  rdrd{\theta}dz

    we first exchange  x and z , it would be easier for us to solve .

    Therefore , the boundary becomes  z = 9r^2 , z = 9

    and when z = 9 ,  9 = 9 r^2  \implies r=1

    the integral becomes  \int\int\int_{E_1} z rdrd{\theta}dz

     = \int_0^{2\pi} \int_0^1 \int_{9r^2}^{9} zr dzdrd{\theta}

     = 2\pi \int_0^1 r \frac{1}{2} (9^2 - 9^2 r^4 ) ~dr

     = 81\pi \int_0^1 (r - r^5)dr  = 27 \pi
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  3. #3
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    Quote Originally Posted by zpwnchen View Post
    what are the limit of z,y and x to integrate?

    I got
    -sqrt(1-y^2) \leq z \leq sqrt(1-y^2)
    -1 \leq y \leq 1
    0 \leq x \leq 9

    help me


    It should be

     \int_{-1}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{9y^2 + 9z^2}^{9} x dxdydz
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