1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    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

    User Avatar
    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



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