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Thread: 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 $\displaystyle z= 18-2x^2-2y^2$ and above the xy-plane

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

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

    (this is the part I'm not sure off) coveting to polar form yields $\displaystyle \int_0^{2\pi} \int_0^{\sqrt{3}}(18-2r^2)r \ dr \ d \theta$
    Last edited by lllll; Jan 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 $\displaystyle z= 18-2x^2-2y^2$ and above the xy-plane

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

    so the paraboloid intersects the xy-plane $\displaystyle -3 \leq x \leq 3$ and $\displaystyle -3 \leq y \leq 3$ 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 $\displaystyle \iint_D 18-2x^2-2y^2 \ dA$

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

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