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Math Help - volume of paraboloid

  1. #1
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    volume of paraboloid

    find the volume of the paraboloid z=x^2+y^2 below the plane z=4.

    i am stuck on setting up the integral..can any one help???
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chris25 View Post
    find the volume of the paraboloid z=x^2+y^2 below the plane z=4.

    i am stuck on setting up the integral..can any one help???
    Let the cross section of the paraboloid at z=4 be our region R.

    Thus, the x and y limits of integration become pretty obvious: \left\{R: \ -2\leq x \leq 2, \ -2\leq y \leq 2\right\}.


    (this is the region R)

    Thus, our volume integral would be \int_{-2}^2\int_{-2}^2 x^2+y^2 \,dx\,dy.

    We can also represent this integral in polar coordinates. Note that the limits of integration change!

    This time we see : \left\{R: \ 0\leq r \leq 2, \ 0\leq \theta \leq 2\pi\right\}.

    As a result, the integral becomes \int_0^{2\pi}\int_0^2 r^3 \,dr\,d\theta.

    Hope that this makes sense!

    --Chris
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Hey, Chris answered Chris!
    Quote Originally Posted by Chris L T521 View Post
    Let the cross section of the paraboloid at z=4 be our region R.

    Thus, the x and y limits of integration become pretty obvious: \left\{R: \ -2\leq x \leq 2, \ -2\leq y \leq 2\right\}.


    (this is the region R)

    Thus, our volume integral would be \int_{-2}^2\int_{-2}^2 x^2+y^2 \,dx\,dy.

    We can also represent this integral in polar coordinates. Note that the limits of integration change!

    This time we see : \left\{R: \ 0\leq r \leq 2, \ 0\leq \theta \leq 2\pi\right\}.

    As a result, the integral becomes \int_0^{2\pi}\int_0^2 r^3 \,dr\,d\theta.

    Hope that this makes sense!

    --Chris
    actually, wouldn't the volume be given by V = \int _{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_{x^2 + y^2}^4 ~dzdydx= \int _{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \bigg[4 - (x^2 + y^2) \bigg]~dydx

    which yields \int_0^{2 \pi} \int_0^2 (4r - r^3)~dr d \theta ?

    ...or am i again?
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