1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

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
    $\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$

    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


    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


    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


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook