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Homework Help: Integralx^3sqrt(4-9x^2)dx with limits of integration at 2/3, 0 (trig subst)

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Integralx^3sqrt(4-9x^2)dx with limits of integration at 2/3, 0 (trig subst)

    2. Relevant equations



    3. The attempt at a solution
    x=sqrt(4/9)sin theta
    dx=sqrt(4/9)cosine theta

    sqrt(4-9x^2)
    sqrt(4-9*4/9sin^2theta)
    sqrt(4-4sin^2theta)
    sqrt(4(1-sin^2theta)
    sqrt(4cos^2theta)
    2costheta

    Integralx^3*2cos theta dtheta
    Integral sin^3*2cos theta dtheta
    Integral (cos^2theta-1)*sin theta*2cos theta dtheta
    Integral (2cos^3 theta*sin theta dtheta)-Integral (2cos theta sin theta dtheta)
    u=cos theta
    du=sin theta dtheta
    2u^4/4du-2u^2/2du
    (2cos^4 theta/4)-(2cos^2 theta/2)

    since x=sqrt(4/9)sin theta
    x=opposite
    sqrt(4/9)=hypoteneuse
    and sqrt(4-9x^2)=adjacent
    so 1/2sqrt(4-9x^2)^4-2/2sqrt(4-9x^2)^2+C
    then plug in the numbers.

    The number that I got was different from the books. Help is appreciated
    since this problem is driving me crazy. Thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 19, 2007 #2
    Remember to keep your coefficients well marked up. =P

    I haven't finished the problem yet, but I can already tell you did that wrong. Go back and do them long hand:

    u = a*sin(theta)
    du = a*cos(theta) d(theta)


    EDIT:
    The answer I got was .1333333 repeating. Was that right? If it is, then check over your beginning again for coefficient mistakes.
     
    Last edited: Jul 19, 2007
  4. Jul 19, 2007 #3

    cepheid

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    Gold Member

    Okay, let's go to the step where you've just carried out the trig substitution for x in one of the factors, but not the other. You have

    [tex] \int_0^{\frac{2}{3}} 2x^3 \cos \theta \, dx [/tex]

    My question is, how come you wrote [itex] d\theta [/itex] instead of [itex] dx [/itex], without first converting the [itex] dx [/itex] into [itex] d\theta [/itex] properly?

    Also, how come when you made the trig substitution for [itex] x^3 [/itex], you got

    [tex] \sin^3 \theta [/tex]

    instead of

    [tex] \left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta [/tex]

    ?
     
  5. Jul 19, 2007 #4
     
  6. Jul 19, 2007 #5
    Never mind him, was my answer right?


    (.133333333)
     
  7. Jul 19, 2007 #6
    The answer is 64/1215.
     
  8. Jul 19, 2007 #7

    VietDao29

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    Homework Helper

    Omg, no LaTeX. My eyes are hurting. :cry: :cry: :cry:
    You should try to learn LaTeX, it's extremely convenient.
    You mean this: [tex]\int_0 ^ \frac{2}{3} \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx[/tex]. Right?

    Well, sqrt(4/9) is actually 2/3. You can rewritten the whole thing as:
    x = (2/3) sin theta
    dx = (2/3) cos theta d(theta)

    So far, so good. :)

    This is where you went wrong.
    When chaging dx to d(theta), and x to sin theta, and cos theta you forget the factors.

    x = (2/3) sin theta
    dx = (2/3) cos theta d(theta)

    You accidentally dropped out the 2/3 factor.

    Well, you should re-do this part. :)

    -----------------------------------

    There's one more convenient way to tackle this problem, i.e to use u-substitution. x3, can be splitted into x2 x

    Letting u = x2

    The whole thing become:

    [tex]\int \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx = \frac{1}{2} \int \ u \sqrt{4 - 9u} \ du[/tex]

    Another u-substitution will do it. Can you go from here? :)
     
  9. Jul 20, 2007 #8
    Wow, thanks VietDao, but I'm gonna see if I could get it done the traditional way to see where I went wrong barring the stupid mistake..........

    Integralx^3*2cos theta dtheta
    Integral (2/3)sin^3*2cos theta dtheta
    Integral 2/3(cos^2theta-1)*sin theta*2cos theta dtheta
    Integral 4/3cos^3theta*sin theta d theta-Integral 4/3 cos theta sin theta d theta
    u=cos theta
    du=sin theta d theta

    4u^4/12-4u^2/6
    4cos^4 theta/12-4cos^2 theta/6+C
     
  10. Jul 20, 2007 #9

    VietDao29

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    Whoops, it's still wrong =.=" You've messed up the coefficients.

    Ok, let's do it step by step then.
    Since x = (2/3) sin(theta)
    x3 = (2/3)3 sin3(theta)
    dx = (2/3) cos(theta) d(theta)

    Change all x to theta, we have:
    [tex]\int \left( \frac{2}{3} \sin ( \theta ) \right) ^ 3 \sqrt{4 - 9 \times \frac{4}{9} \sin ^ 2 \theta} \left( \frac{2}{3} \cos \theta \ d ( \theta ) \right) = \int \frac{8}{27} \sin ^ 3 ( \theta ) \left( 2 \cos \theta \right) \left( \frac{2}{3} \cos \theta \ d ( \theta ) \right)[/tex]

    Now, hopefully, you can take it from here, right? :)
     
  11. Jul 20, 2007 #10
    Integral 32/81 (1-cos^2theta)*cos theta*sin theta d theta
    Integral 32/81 cos theta *sin theta d theta-Integral 32/81 cos^3 theta * sin theta d theta
    u=cos theta
    du=sin theta
    32u^2/162-32u^4/324
    32cos^2 theta/162-32cos^4 theta/324
     
  12. Jul 21, 2007 #11

    VietDao29

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    Well, the first line is incorrect. >"<
    Actually, it is cos2(theta) instead of cos(theta). Common, try again one more time, man, you are very very close to the answer. Just correct some minor mistakes, and it's done. :)

    Oh, and by the way. If u = cos(theta), then du would be: du = -sin(theta)d(theta)
     
    Last edited: Jul 21, 2007
  13. Jul 21, 2007 #12
    Then I'll just simply change the first line to
    Integral 32/81 cos^2 theta sin theta d theta- Integral 32/81 cos^4 theta sin theta d theta
    u=cos theta
    du= -sin theta d theta
    32u^3/243-32u^5/405
    32 cos^3 theta/243-32 cos^5 theta/405:wink:
     
  14. Jul 21, 2007 #13

    VietDao29

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    Well, you still got the wrong signs, though. du = -sin(theta)d(theta). There's a minus sign in front of it. :)

    Ok, well, yeah, just change the signs, and you'll arrive at the final answer. :) ^.^
     
  15. Jul 21, 2007 #14
    32cos^3 theta/243+32cos^5 theta/405
     
  16. Jul 21, 2007 #15

    VietDao29

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    Homework Helper

    No... It's still wrong. :cry: :cry: :cry: :cry:
    -(a - b) = -a + b, not a + b

    You should flip both signs.
    -32 cos^3 theta/243+32 cos^5 theta/405

    Well, you should do some more manipulations, it's good for some problem like this, i.e, require a lot of calculation.

    Btw, congratulations. Finally, you got it :)
     
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