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Math Help - Volume under solid and polar integration

  1. #1
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    Volume under solid and polar integration

    I'm having some trouble finding the area under z= 18-2x^2-2y^2 and above the xy-plane

    starting off z= 18-2(x+3)(y-3) and z= 18-2(y+3)(x-3) so the paraboloid intersects the xy-plane -3 \leq x \leq 3 and -3 \leq y \leq 3

    taking the integral would wield \iint_D 18-2x^2-2y^2 \ dA

    (this is the part I'm not sure off) coveting to polar form yields \int_0^{2\pi} \int_0^{\sqrt{3}}(18-2r^2)r \ dr \ d \theta
    Last edited by lllll; January 12th 2009 at 06:10 PM.
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  2. #2
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    Quote Originally Posted by lllll View Post
    I'm having some trouble finding the area under z= 18-2x^2-2y^2 and above the xy-plane

    starting off z= 18-2(x+3)(y-3) and z= 18-2(y+3)(x-3)

    so the paraboloid intersects the xy-plane -3 \leq x \leq 3 and -3 \leq y \leq 3 Mr F says: No. This implies that the region of integration in the xy-plane is a square and, as we both know, it's not. It's a disk (see below).

    taking the integral would wield \iint_D 18-2x^2-2y^2 \ dA

    (this is the part I'm not sure off) coveting to polar form yields \int_0^{2\pi} \int_0^{{\color{red}3}}(18-2r^2)r \ dr \ d \theta
    See the correction (in red). The area of the xy-plane you're integrating over is the disk x^2 + y^2 \leq 9. The radius of this disk is 3, not \sqrt{3}.

    To get the region, solve z = 0 \Rightarrow 0= 18-2x^2-2y^2.
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