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Trig problem involving parametric equations

  1. Sep 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Let R be the region in the 1st quadrant in the region enclosed by [tex]x=2cos(\theta)[/tex] and [tex]y=sin(2\theta)[/tex] Suppose R is rotated around the x-axis.

    Find the volume of the resulting solid.


    2. Relevant equations

    The formula for the solid of revolution is:

    [tex]V= \pi\int y^2 dx = \pi\int y^2 f(x)' dx [/tex]

    I've included a picture of the graph below.
    [​IMG]

    3. The attempt at a solution

    I plugged in the numbers into the formula and factored out the constants, but am now stuck with this crazy integral, did I do something wrong? I'm doing an introductory integral course, should I be able to solve these kind of integrals you think?

    [tex] V= -2\pi\int Sin^2(2\theta) Sin(\theta) d\theta [/tex]
     
  2. jcsd
  3. Sep 5, 2008 #2

    Defennder

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    You need to figure out the limits of theta as it traces out the path it parametrises. It's given in the diagram. Note that the path begins at (2,0) and ends at the same point, but don't take 0 to 2pi as your limits of theta because otherwise you get zero since it returns to back where you started. Instead, you have the endpoint as the other end of the graph at (-2,0). Use the difference in theta to get both the limits of integration.

    But this is unnecessarily complicated if you ask me. If you stick to using x as your integration variable, then all you have to do is to express y in terms of x and integrate over the range of x of the graph. It's easier to calculate this way.
     
  4. Sep 8, 2008 #3
    Alright, but this integral doesnt seem any easier to integrate:

    [tex]
    V= \int Sin(ArcCos(x/2)) d\theta
    [/tex]

    I was thinking, as this question is part of bigger exercise where I calculated the area to begin with then perhaps this help me in any way to find the volume?
     
    Last edited: Sep 8, 2008
  5. Sep 8, 2008 #4

    Defennder

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    First you don't have to write it as [tex]\sin (\cos^{-1} \frac{x}{2})[/tex]. You have [tex]\cos \theta = \frac{x}{2}[/tex] and you can easily come up with an expression for [tex]\sin \theta[/tex] from drawing the right-angle triangle for this.

    Secondly, if you're performing the integration with respect to x, your integral dummy variable should be [tex]dx[/tex]. There's also a square which you didn't include.

    Is that supposed to be the area enclosed by the curve? I may be mistaken, but I just don't see how the area of a closed loop is related to the volume it generates by revolving it about an axis, especially since the volume differs when the axis of rotation is changed. To take an example, the area of a circle is pi(r^2). How can you get the volume of the sphere generated from the expression of area alone without doing any integration?
     
    Last edited: Sep 8, 2008
  6. Sep 8, 2008 #5
    Trigonometry is seriously not my strong point so correct me if I am wrong, but since there's two unknowns, [tex]\theta[/tex] and X I am not sure how I can find an expression for [tex]Sin \theta[/tex]

    In any case, furthermore, say I did find an expression for [tex]Sin \theta[/tex], I'm wouldnt really know how I could plug that information into [tex]y = Sin (2\theta)[/tex]. I am thrown off by the 2...
     
  7. Sep 8, 2008 #6

    Defennder

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    Use the double angle formula for sine to reduce sin 2theta into sine and cosin expressions. You have x = 2cos(theta). So that means cos(theta) = x/2. So draw the right-angle triangle and mark one corner as theta, and the corresponding triangle lengths as x and 2. Now use pythagoras theorem to get the unknown side. Then reading from this triangle, what is sin(theta) ?
     
  8. Sep 17, 2008 #7
    Okay, so I've got it up like on the diagram.

    [​IMG]

    Doing my high school math right, [tex]Sin \theta = \frac{\sqrt{4-x^2}}{2}[/tex]

    [tex]y = Sin(2\theta) = 2 Sin(\theta) Cos(\theta)[/tex]

    So plugging in all I know, after simplification I get: [tex]y = \frac{x\sqrt{4-x^2}}{2} [/tex]

    And then I take the integral of this? Am I on the right track?

    Edit: Correction, the answer should be, [tex]y = x\sqrt{4-x^2}} [/tex], but I can't figure out why the \frac{1/2} fraction cancels out.
     
    Last edited: Sep 17, 2008
  9. Sep 17, 2008 #8

    Defennder

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    Why does 1/2 cancel out?
     
  10. Sep 17, 2008 #9
    Whoopsie, you're absolutely right, 1/2 stays!

    I solved the problem!

    For those interested in the solution, taking the integral of the above expression between x=0 and x=2 gives you the answer 1.3334 which is indeed what I was looking for :)

    Thanks for your help, Defennder!
     
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