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Definite integral of a quarter circle

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the definite integral of a quarter circle.

    2. Relevant equations
    x^2 +y^2=10




    3. The attempt at a solution
    x^2=10-y^2
    x=sqrt(10-y^2)

    ∫ sqrt(10-y^2)dy from 0 to sqrt(10)

    I'm not sure what to do here.
     
  2. jcsd
  3. Dec 11, 2012 #2

    Mute

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    Homework Helper

    Well, it's a circle, so you can probably guess that a factor of ##\pi## should pop out somehow. What kind of substitution do you think that fact might suggest you should use?
     
  4. Dec 11, 2012 #3
    I guess I should use a trig substitution. 10cos(θ)^2 +10sin(θ)^2=10
    √(10) cos(θ) = √(10-10sin(θ)
    y=√(10)sin(θ)
    ∫√(10-(10)sin(θ)) dy
    dy=√(10)cos(θ)dθ
    ∫√(10-(10)sin(θ)) √(10) cos(θ) dθ
    ∫√(10) cos(θ) √(10) cos(θ) dθ
    arcsin(t/sqrt(10))=θ
    10 ∫cos(θ)^2 dθ

    how do i evaluate that last integral?
     
    Last edited: Dec 11, 2012
  5. Dec 11, 2012 #4

    Mute

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    There's an identity that relates ##\cos^2\theta## to ##\cos(2\theta)##. Do you know it? If not, you can derive it. Consider the double angle formula for cos:

    $$\cos(A+B) = \cos A \cos B - \sin A \sin B$$

    Pick A and B appropriately and then use Pythagoras' theorem to derive the identity, then plug that into your integral.
     
  6. Dec 11, 2012 #5
    cos(2x)=cos^2(x)-1 +cos^2(x)
    cos(2x)=2cos^2(x)-1
    cos(2x)/2 +1/2

    5∫ 1 dθ + ∫ cos(2θ) dθ

    5θ + sin(2θ)/2 from 0 to pi/2

    5pi/2 ?

    Oh man that was not fun. :/ Thanks for your help. I REALLY appreciate it.
     
  7. Dec 11, 2012 #6

    Mute

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    Yep, that's the answer! It's easy to check, too: since it's a quarter of a circle of radius ##\sqrt{10}##, you expect the answer to be ##\pi (\sqrt{10})^2/4 = 5\pi/4##, which is what you got.

    There's another way to do this which is simpler, but you have to start it off a bit differently than you did, and I didn't know if you've studied polar coordinates or double integrals yet.

    You integral is basically

    $$\int_0^{\sqrt{10}} dy \int_0^{\sqrt{10-y^2}}dx;$$
    doing the x integral gives you the integral you started with. However, converting to polar coordinates ##x = r\cos\theta##, ##y=r\sin\theta##, ##dxdy \rightarrow d\theta dr r## turns the integral into

    $$\int_0^{\pi/2} d\theta \int_0^{\sqrt{10}} dr r,$$

    which is much easier to do.
     
  8. Dec 12, 2012 #7

    HallsofIvy

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    Odd. I would have considered that a lot of fun and very satisfying!
     
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