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Thread: Volume integral

  1. #1
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    Volume integral

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

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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 $\displaystyle x^2 + y^2 = z^2~ (z \ge 0)$ and the paraboloid $\displaystyle z =2x^2 + 2y^2$.

    ------

    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 $\displaystyle z=0, \frac{1}{2}.$ for $\displaystyle 0 \leq z \leq \frac{1}{2}$ the cone is the top surface. so the volume is: $\displaystyle V= \int \int_R [\sqrt{x^2+y^2} - 2(x^2+y^2)] dA,$ where $\displaystyle R=\{(x,y): \ x^2+y^2 \leq \frac{1}{4} \}.$

    using polar coordinates we get: $\displaystyle 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:

    $\displaystyle x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?

    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. $\displaystyle z_1 - z_2$?
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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:

    $\displaystyle x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?

    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. $\displaystyle z_1 - z_2$?
    to find the intersection: $\displaystyle z^2=x^2+y^2=\frac{z}{2}.$ thus $\displaystyle 2z^2=z$ and hence $\displaystyle z=0, \frac{1}{2}.$ to answer your second question, recall that the volume of the region $\displaystyle D$ bounded by the two surfaces:

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