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Math Help - Using integral to find volume problem

  1. #1
    iva
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    Using integral to find volume problem

    Hi there,

    I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

    Calculate volume of D where D is the region in R^3 enclosed by the surfaces
    z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

    So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

    I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
    So I integrated over

    theta = 0 to 2pi ( ie all the way around)
    r= sqrt(3) by equating the 2 parabaoids to find the intesection
    z integrated over 3-r^2 and r^2-3.

    The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
    So i decided to integrate in 2 identical parts split at z=0.
    I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

    (if this is impossible to read I'll reattempt the Latex)

    Thank you
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by iva View Post
    Hi there,

    I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

    Calculate volume of D where D is the region in R^3 enclosed by the surfaces
    z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

    So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

    I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
    So I integrated over

    theta = 0 to 2pi ( ie all the way around)
    r= sqrt(3) by equating the 2 parabaoids to find the intesection
    z integrated over 3-r^2 and r^2-3.

    The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
    So i decided to integrate in 2 identical parts split at z=0.
    I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

    (if this is impossible to read I'll reattempt the Latex)

    Thank you
    Your idea is sound.

    \int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{r^2-3}^{3-r^2}rdzdrd\theta=2\int_{0}^{2\pi}\int_{0}^{\sqrt{3  }}(3r-r^3)drd\theta=2\left( \int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\sqrt  {3}} (3r-r^3)dr\right)
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by iva View Post
    Hi there,

    I have an exercise I couldn't solve, have the answer which i couldn't get to, any guidance will be appreciated:

    Calculate volume of D where D is the region in R^3 enclosed by the surfaces
    z=3-x^2-y^2 and z=x^2+y^2 -3 (Sorry Latex was giving me errors)

    So this looks like 2 identical parabaloids one startign at z=3 going down and one starting at z=-3 going up.

    I used a triple integral with cylindrical coordinates ( if theres another better way let me know)
    So I integrated over

    theta = 0 to 2pi ( ie all the way around)
    r= sqrt(3) by equating the 2 parabaoids to find the intesection
    z integrated over 3-r^2 and r^2-3.

    The first issue here is that when i do the z integration i come out with zero (what did I do wrong here)
    So i decided to integrate in 2 identical parts split at z=0.
    I followed this through but didn't get the books answer of 9pi. i got 24root(3)/5 for one half, how does the book get to 9 pi? At which step did I go wrong?

    (if this is impossible to read I'll reattempt the Latex)

    Thank you
    V=\int_0^{2\pi}\int_0^{\sqrt{3}}(z_{top}-z_{bottom})rdrd\theta

    = \int_0^{2\pi}\int_0^{\sqrt{3}}(6-2r^2)rdrd\theta

    = \int_0^{2\pi}\int_0^{\sqrt{3}}(6r-2r^3)drd\theta

    = \int_0^{2\pi}\left[3r^2-\frac{r^4}{2}\right]_0^{\sqrt{3}}d\theta

    = \int_0^{2\pi}(3\times 3-\frac{9}{2})d\theta

    = \int_0^{2\pi}\frac{9}{2}d\theta

    = \frac{9}{2}\times 2\pi

    = 9\pi
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  4. #4
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    the "z integration" is NOT 0, you should wind up with r((3-r^2) - (-3+r^2)) = 6r - 2r^3.

    integrating that with respect to r from 0 to √3, will yield 3r^2 - \frac{r^4}{2}, evaluated at √3 which gives you 9/2.

    integrating the constant function 9/2 over θ = 0 to θ = 2\pi, gives 9\pi, as your textbook indicates.
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