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Bad surface integral

  1. Apr 7, 2006 #1
    [tex]\int x {(C - x^{2/3})}^{3/2} dx[/tex]

    Any ideas?
     
    Last edited: Apr 7, 2006
  2. jcsd
  3. Apr 7, 2006 #2

    VietDao29

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    [tex]\int x \sqrt{ \left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx[/tex]
    To solve integrals with square roots like that, one can start off their problem by changing the variable x to some variable t such that:
    [tex]C - \sqrt[3]{x ^ 2} = C \cos ^ 2 t[/tex]
    So, let:
    [tex]x = \sqrt{C ^ 3} \sin ^ 3 t \quad \quad \quad t \in \left[ -\frac{\pi}{2} ; \ \frac{\pi}{2} \right][/tex]
    [tex]\Rightarrow dx = 3 \sqrt{C ^ 3} \cos t \sin ^ 2 t dt[/tex].
    Substitute all to your integral yields:
    [tex]\int x \sqrt{\left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx = 3 C ^ 3 \int \sin ^ 5 t \cos t \sqrt{ \left( C - \sqrt[3]{C ^ 3 \sin ^ 6 t} \right) ^ 3} dt = 3 C ^ 3 \int \sin ^ 5 t \cos t \sqrt{ \left( C - C \sin ^ 2 t} \right) ^ 3} dt[/tex]
    [tex]= 3 \sqrt{C ^ 9} \int \sin ^ 5 t \cos ^ 4 t dt[/tex]
    You can go from here, right? :)
     
    Last edited: Apr 7, 2006
  4. Apr 7, 2006 #3
    Yup, thanks! :)
    But isn't this substitution very subtle? Am I an oracle to guess it instantenously in an exam!?
     
  5. Apr 7, 2006 #4
    Well it may appear ingenious, but really [itex]sin^2t + cos^2t=1[/itex] so the idea of substituting some function of [itex]sin t[/itex] is going to be a good idea. Once you have that idea, you can work out what powers you need &c.
     
  6. Apr 8, 2006 #5

    VietDao29

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    Nah, you don't need to be an oracle to know it.
    To solve square root, there are some common ways:
    The first way is to use an u substitution:
    u2 = everything inside the square root, so that when you squre root it, it'll become u. In your problem:
    [tex]u ^ 2 = C - \sqrt[3]{x ^ 2}[/tex].
    Then [tex]2u du = -\frac{2}{3 \sqrt[3]{x}} dx \Rightarrow dx = 3u \sqrt[3]{x}[/tex].
    Substitute all to your integral, and it will become:
    [tex]\int x \sqrt{ \left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx = 3 \int u ^ 4 \sqrt[3]{x ^ 4} du = 3 \int u ^ 4 (C - u ^ 2) ^ 2 du[/tex]
    Try it to see if you can arrive at the same answer as you previously got.
    ------------------
    The second way is to use trig substitution. Change x to u such that everything inside the square root become something * cos2u, so that again when you square root it, it will be come: [tex]\sqrt{\mbox{something}} \times \cos u[/tex].
    So let:
    [tex]C - \sqrt[3]{x ^ 2} = C \cos ^ 2 t[/tex]
    [tex]\Leftrightarrow \sqrt[3]{x ^ 2} = C \sin ^ 2 t[/tex]
    [tex]\Leftrightarrow x = \sqrt{C ^ 3} \sin ^ 3 t[/tex]
    [tex]\Rightarrow dx = 3 \sqrt{C ^ 3} \cos t \sin ^ 2 t dt[/tex]
    ...
    Can you get this? :)
     
  7. Apr 8, 2006 #6

    arildno

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    Dearly Missed

    Alternatively, you may choose:
    [tex]y=x^{\frac{1}{3}}\to{dx}=3y^{2}dy[/tex]
    And, hence:
    [tex]\int{x}(C-x^{\frac{2}{3}})^{\frac{3}{2}}dx=\int{3y^{5}}(C-y^{2})^{\frac{3}{2}}dy[/tex]

    This is solved nicely by repeated integrations by parts.
     
  8. Apr 10, 2006 #7
    even better...

    Let [tex]u = \sqrt{C-x^{\frac{2}{3}}[/tex]


    Then [tex]x = (C-u^2)^{\frac{3}{2}} \Rightarrow[/tex]
    ...

    [tex]dx = \frac{3}{2}(C-u^2)^{\frac{1}{2}}(-2)udu\Rightarrow[/tex]

    ...

    [tex]dx = -3(C-u^2)^{\frac{1}{2}}udu[/tex]
    ...
    Then: [tex]\int x(C-x^{\frac{2}{3}})^{\frac{3}{2}}dx = \int[ (C-u^2)^{\frac{3}{2}})(u^3)(-3(C-u^2)^{\frac{1}{2}}udu)][/tex]

    Put the two [tex] (C-u^2)[/tex] terms together and simplifying:

    [tex] = -3\int (C-u^2)^2u^4du \Rightarrow -3\int(c^2 - 2Cu^2 + u^4)(u^4)du[/tex]

    ...
    [tex]=-3\int(u^4c^2 - 2u^6C + u^8)du[/tex]
    ...
    [tex] =-3[C^2\int u^4du - 2C\int u^6 du + \int u^8 du][/tex]

    :)
     
  9. Apr 10, 2006 #8

    VietDao29

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    Have you looked closely at the fifth post of this thread, daveyp225?? :rolleyes:
     
  10. Apr 10, 2006 #9
    Evidently not.
     
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