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Math Help - triple integral in cylindrical coordinates

  1. #1
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    Question triple integral in cylindrical coordinates

    Convert and evaluate the integral of (from -1 to 1) the integral of (0 to sqrt(1-y^2) ) the integral of (from x^2+y^2 to sqrt(x^2 +y^2) ) xyz dz dx dy into an integral in cylindrical coordinates.

    Please do show me the answer of this question if possible. Thank you very much.
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  2. #2
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    Quote Originally Posted by kittycat View Post
    Convert and evaluate the integral of (from -1 to 1) the integral of (0 to sqrt(1-y^2) ) the integral of (from x^2+y^2 to sqrt(x^2 +y^2) ) xyz dz dx dy into an integral in cylindrical coordinates.

    Please do show me the answer of this question if possible. Thank you very much.
    \int_{-1}^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}}xyz \ dz \ dx \ dy.

    Look at the area over which we are integrating expressed by 0\leq x\leq \sqrt{1-y^2} \mbox{ and }-1\leq y \leq 1.
    You should realize that this is a semi-circle in the right half plane. Which is expressed in polar form as -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \mbox{ and }0\leq r\leq 1.


    Using r=\sqrt{x^2+y^2} the integral becomes,
    \int_{-\pi/2}^{\pi/2} \int_0^1 \int_{r^2}^r (r\cos \theta)(r\sin \theta) z r \ dz \ dr \ d\theta
    Use double angle identity,
    \frac{1}{2}\int_{-\pi/2}^{\pi/2}\int_0^1 \int_{r^2}^r r^3 \sin 2\theta z \ dz \ dr \ d\theta.
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  3. #3
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    Hi perfecthacker,

    Thank you very much.
    Is the answer 1/48?
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