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    calc help

    Find the volume of the solid enclosed by the paraboloids z=16(x^2+y^2) and z=18-16(x^2+y^2)
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    Quote Originally Posted by karnold9 View Post
    Find the volume of the solid enclosed by the paraboloids z=16(x^2+y^2) and z=18-16(x^2+y^2)
    The top surface is z=18-16(x^2+y^2) and lower surface is z=16(x^2+y^2). When these curves intersect we have 18 - 16(x^2+y^2) = 16(x^2+y^2) thus x^2+y^2 = \tfrac{9}{16}. This means the area over we are integrating is disk x^2+y^2\leq \left( \tfrac{3}{4} \right)^2.
    The required integral is,
    \iint_A 18 - 16(x^2+y^2) - 16(x^2+y^2) dA = \int_0^{2\pi} \int_0^{3/4} (18 - 32r^2)r ~ dr ~ d\theta.
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