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Homework Help: Trigonometric Inegrals: Volume help

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Let R be the region enclosed by the curves y=6sin2x and y=6 over the segment [ π/2,3π/2]. Find the volume of the solid that is obtained by rotating R about the line y=−1.1.


    2. Relevant equations
    [tex]
    $\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$
    [/tex]

    3. The attempt at a solution
    [tex] $\displaystyle \Large pi * \int _{pi/2}^{3pi/2} x((-1.1+6)^2 - (-1.1+6sin^2x)^2)dx$ [/tex]
    ==> [tex] $\displaystyle \Large pi * \int _{pi/2}^{3pi/2} 24.01x - (-1.1+6sin^2x)^2 xdx$ [/tex]

    ==> [tex] $\displaystyle \Large pi [* \int _{pi/2}^{3pi/2} 24.01x dx - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$ [/tex]
    ==> [tex] $\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$ [/tex]
    ==> [tex] $\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} 36xsin^4x - 13.2xsin^2x + 1.21x dx ]$ [/tex]
    ==> [tex] $\displaystyle \Large pi [* (236.97 + C) - (36\int _{pi/2}^{3pi/2} xsin^4xdx - 13.2\int _{pi/2}^{3pi/2}xsin^2xdx + 1.21\int _{pi/2}^{3pi/2}xdx)]$ [/tex]

    is this right so far?
     
    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 13, 2010 #2

    Mark44

    Staff: Mentor

    No, it looks like you're starting off on the wrong foot here. The integral above isn't right for a couple of reasons. That first factor of x in the integral doesn't belong there, and you should be subtracting -1.1 (i.e., adding + 1.1), instead of just subtracting 1.1.
    You're obviously using washers, so here's what your integral should look like.
    [tex] \pi * \int _{\pi/2}^{3\pi/2} ((6 -(-1.1))^2 - (6sin^2x -(-1.1))^2)dx [/tex]

    In a slightly simpler form this is:
    [tex] \pi * \int _{\pi/2}^{3\pi/2} ( 7.1^2 - (6sin^2x + 1.1)^2)dx [/tex]
    In a typical volume element ("washer"), the volume is [itex]\pi(R^2 - r^2)\Delta x [/itex]. The larger radius R is 6 - (-1.1) = 7.1. The smaller radius r is 6sin2(x) - (-1.1) = 6sin2(x) + 1.1.

     
  4. Feb 13, 2010 #3
    i did exactly what you said and i get 451.7359646 but its wrong..
    [tex]$\displaystyle \Large pi [(50.41x + C) - (36\int _{pi/2}^{3pi/2} sin^4xdx + 13.2\int _{pi/2}^{3pi/2}sin^2xdx + 1.21x+C)]$[/tex]

    == >
    [tex]$\displaystyle \Large pi [158.367 - (36\int _{pi/2}^{3pi/2} sin^4xdx + 13.2\int _{pi/2}^{3pi/2}sin^2xdx + 3.8013)]$[/tex]

    then i found the antiderivative of sin4x and sin2x and got the final answer of the Volume. what did i mess up on?
     
  5. Feb 13, 2010 #4

    Mark44

    Staff: Mentor

    Your answer is high by about a factor of 15. I get approximately 29.5783. After a little simplification you should have
    [tex]\pi \int_{\pi/2}^{3\pi/2} (49.2 - 36sin^4(x) -13.2sin^2(x))dx[/tex]

    You forgot to subtract 1.21 (=1.1^2) from the 50.41 term, and it looks like you forgot that pi multiplies all of your integrals. Also, you're dealing with a definite integral: you don't need to include the constant of integration.
     
  6. Feb 14, 2010 #5
    how do you get 29.5783?
    i get 256.73 and im not sure if its right or not but im pretty sure i did everything right this time.
    what i forgot to do was to multiply the '-' sign inside the brackets.
    [tex] \pi [49.2x|_{\pi/2}^{3\pi/2} - 36\int_{\pi/2}^{3\pi/2} sin^4(x)dx- 13.2\int_{\pi/2}^{3\pi/2}sin^2(x)dx] [/tex]
    [tex] = \pi [49.2\pi - 36(3x/2 - sin2x + sin(4x)/8)|_{\pi/2}^{3\pi/2} - 13.2/2(x - sin(2x)/2)_{\pi/2}^{3\pi/2}] [/tex]

    [tex] = \pi [154.57 - 48.82 - 24.03] [/tex] [tex] = \pi [81.72] [/tex] = 256.73
     
    Last edited: Feb 14, 2010
  7. Feb 14, 2010 #6

    Mark44

    Staff: Mentor

    Now I'm getting a different answer, 287.205. I'm using a computer algebra system, and I must have entered something wrong to get the lower number I posted earlier.

    In your integral your terms in x are off.
    [itex]\int sin^4(x) dx = 3x/8 - 1/4*sin(2x) + 1/32*sin(4x)+C[/itex]
    [itex]\int sin^2(x) dx = x/2 - 1/4*sin(2x) + C[/itex]

    Notice also that when you substitute in 3pi/2 and pi/2, all of the sine terms are zero.
     
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