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Area of the region

  1. Apr 8, 2005 #1
    Find the area of the region bounded by: [tex]r^2 = 128*cos(2\theta)[/tex]

    [tex]\int 1/2*(\sqrt{128*cos(2\theta)})^2[/tex]

    that's my integral, but i dont know what the bounds are. i tried typing it into my calculator, but i could only find one bound(.78539816) and im not even sure if this is even correct. can someone help me find the bounds?
     
  2. jcsd
  3. Apr 8, 2005 #2

    AKG

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    You know that cosine is periodic in 2pi, so theta can go from 0 to 2pi. Just by looking at it, you should be able to see that it will be a four leaf clover, and that each of the leaves will have the same area, so the total area will be 4 x the area of one. You should also be able to see that, since cosine is zero at pi/2 and 3pi/2, that one of the clover leafs will be from pi/4 to 3pi/4. And don't worry about r being negative, since the leaf created by taking r negative and theta going from pi/4 to 3pi/4 is just the leaf taking r positive going form 5pi/4 to 7pi/4, which you're already counting when taking the area of the one leaf and multiplying it by 4.

    So your answer will be 4 x area of one leaf, where r is positive and theta ranges from pi/4 to 3pi/4.
     
  4. Apr 8, 2005 #3

    ehild

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    r^2 can not be negative, so

    [tex]-\pi/4\le \theta \le \pi/4 [/tex] and [tex]3\pi/4\le \theta \le 5\pi/4[/tex]

    are your boundaries.

    ehild
     
    Last edited: Apr 8, 2005
  5. Apr 8, 2005 #4

    AKG

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    How could you possibly get -64? I don't remember the formula for finding the area bounded by a curve given in polar co-ordinates, but if the formula you're using is correct, your integrand is strictly positive (it's a squared number (positive) times 1/2 (positive)).

    Oh, I can't believe I didn't notice this. Something is wrong (maybe) since cosine will take on negative values on that region, but r² is positive, so I suppose it is implied that theta will only range over values where the equation makes sense. Cosine is non-negative from 3pi/2 to pi/2, so integrate from 3pi/4 to pi/4 (and not the other way around. Now, I'm guessing that if you're given r² = 128*cos(theta), maybe they want you to take the positive and negative values of r, so you'll get a 2-leafed clover. The area you need is then 2 x the area from 3pi/4 to pi/4 (of course, use 5pi/4, not pi/4), and that area is just 1/2r² d(theta), and r² is given to you, you don't have to write it as the square of a square root. You need to compute:

    [tex]2\int _{\frac{3\pi }{4}} ^{\frac{5\pi }{4}} \frac{1}{2}r^2\, d\theta[/tex]

    [tex]= \int _{\frac{3\pi }{4}} ^{\frac{5\pi }{4}} 128\cos (2\theta)\, d\theta[/tex]

    [tex]= 64\sin (2\theta)|_ {\theta = \frac{3\pi }{4}} ^{\theta = \frac{5\pi }{4}}[/tex]

    [tex]= 64\left (\sin \left (\frac{pi}{2}\right ) - \sin \left (\frac{3\pi }{2}\right )\right )[/tex]

    [tex] = 128[/tex]
     
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