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Math Help - Volume integral

  1. #1
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    Volume integral

    Find the volume of the region bounded by the cone x^2 + y^2 = z^2~ (z \ge 0) and the paraboloid z =2x^2 + 2y^2.

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    I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.



    Would I do this in rect-coords or cylindrical or what?
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Find the volume of the region bounded by the cone x^2 + y^2 = z^2~ (z \ge 0) and the paraboloid z =2x^2 + 2y^2.

    ------

    I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.



    Would I do this in rect-coords or cylindrical or what?
    two surfaces intersect at z=0, \frac{1}{2}. for 0 \leq z \leq \frac{1}{2} the cone is the top surface. so the volume is: V= \int \int_R [\sqrt{x^2+y^2} - 2(x^2+y^2)] dA, where R=\{(x,y): \ x^2+y^2 \leq \frac{1}{4} \}.

    using polar coordinates we get: V=\int_0^{2 \pi} \int_0^{\frac{1}{2}}(r^2-2r^3) \ dr d \theta = \frac{\pi}{48}.
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  3. #3
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    Thanks, but I have some questions:

    How did you find their intersection? Did you solve the quartic polynomial:

    x^2+y^2 = 4x^4+8x^2y^2+4y^4 ?

    Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...

    How did you determine that the integrand should be the difference between the surface functions, I.e. z_1 - z_2?
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    Thanks, but I have some questions:

    How did you find their intersection? Did you solve the quartic polynomial:

    x^2+y^2 = 4x^4+8x^2y^2+4y^4 ?

    Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...

    How did you determine that the integrand should be the difference between the surface functions, I.e. z_1 - z_2?
    to find the intersection: z^2=x^2+y^2=\frac{z}{2}. thus 2z^2=z and hence z=0, \frac{1}{2}. to answer your second question, recall that the volume of the region D bounded by the two surfaces:

    z=f(x,y), \ z=g(x,y), with f(x,y) \leq g(x,y), \ \forall (x,y) \in R, is \int \int_R (g(x,y) - f(x,y)) dA, where R is the image of D on the xy plane.
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  5. #5
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    Ah yes! Thank you very much!
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