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Thread: Green's Theorem

  1. #1
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    Green's Theorem

    how can greens theorem be verified for the region R defined by $\displaystyle (x^2 + y^2 \leq 1), (x + y \geq 0), (x - y \geq 0)$ .... P(x,y) = xy, Q(x,y) = $\displaystyle x^2$
    > okay i know $\displaystyle \int_C Pdx + Qdy = \int\int \left(\frac{dQ}{dx} - \frac{dp}{dy}\right) dA$

    so: $\displaystyle \int_C xy dx + x^2dy = \int\int_D \left(2x - x\right) dy dx$

    but i can't figure out the limits for the double integral: $\displaystyle \int\int_D \left(2x - x\right) dy $... I know they can be found by those inequalities but i'm reachin a dead end.. any suggestions and working out please?

    also what is the region of integration for the left hand side please?
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  2. #2
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    Look at picture below.

    This can be expressed in polar coordinates as:
    $\displaystyle -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \mbox{ and }0\leq r\leq 1$
    Attached Thumbnails Attached Thumbnails Green's Theorem-picture29.gif  
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    This can be expressed in polar coordinates as:
    $\displaystyle -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \mbox{ and }0\leq r\leq 1$
    i understand the limits for "r"... but how do you explain the ranges you have chosen for theta (i.e. pi/4 to -pi/4)??

    anyhow i've calculated it:
    ok so using polar co-ordinates:
    2x - x = x
    in polar terms: x = $\displaystyle rcos\theta$

    so the integral is now:

    $\displaystyle \int_C xy dx + x^2dy = \int^{\pi/4}_{-\pi/4}\int^1_0 \left(r^2cos\theta\right) drd\theta $ = $\displaystyle \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}r^3cos\theta\right]^1_0d\theta $ = $\displaystyle \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}cos\theta\right]d\theta $ = $\displaystyle \left[\frac{1}{3}sin\theta\right]^{\pi/4}_{-\pi/4} $ = $\displaystyle \frac{2}{3}sin(\pi/4)$

    is this correct?

    and now i need to show the other side is equal to 2/3sin(pi/4) right? i.e $\displaystyle \int_C xy dx + x^2dy = 2/3sin(\pi/4)$
    could you give me a hint on how to integrate this one please. thanks mate.
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  4. #4
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    lol cheers. you've done the working for me joanne. i was acutally about to post the same question i.e. how do you show the left hand formula to be equal to 2/3sin(pi/4)
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  5. #5
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    Quote Originally Posted by smoothman View Post
    lol cheers. you've done the working for me joanne. i was acutally about to post the same question i.e. how do you show the left hand formula to be equal to 2/3sin(pi/4)
    You need to use three different line integrals.

    $\displaystyle \bold{r}_1(t) = t\cos \frac{\pi}{4}\bold{i}-t\sin \frac{\pi}{4}\bold{j} \mbox{ for }0\leq t\leq 1$

    $\displaystyle \bold{r}_2(t) = t\cos \frac{\pi}{4}\bold{i}+t\sin \frac{\pi}{4}\bold{j} \mbox{ for }0\leq t\leq 1$

    $\displaystyle \bold{r}_3(t) = \cos t\bold{i} + \sin t \bold{j} \mbox{ for }-\frac{\pi}{4} \leq t\leq \frac{\pi}{4}$
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  6. #6
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    cool thanks. ill work out the line integrals now.

    was my answer correct though? $\displaystyle \frac{2}{3}sin(\pi/4)$
    so now i have to work out that the line integrals also add up to my answer? correct?
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  7. #7
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    here is what i managed to work out so far for the line integrals:

    i know $\displaystyle \int Pdx + Qdy = \int xy dx + \int x^2 dy$ since P=xy and Q=x^2

    for r1(t)
    $\displaystyle
    \bold{r}_1(t) = t\cos \frac{\pi}{4}\bold{i}-t\sin \frac{\pi}{4}\bold{j} \mbox{ for }0\leq t\leq 1
    $

    so:

    $\displaystyle \int_c xy dx + \int_c x^2 dy= \int^1_0 -\frac{1}{2}t^2(sin\frac{\pi}{2})(cos\frac{\pi}{4}) dt$ + $\displaystyle \int^1_0 t^2(cos^2\frac{\pi}{4})(-sin\frac{\pi}{4}) dt
    $

    = $\displaystyle \int^1_0 -\frac{1}{2}t^2(cos\frac{\pi}{4}) dt$ + $\displaystyle \int^1_0 -\frac{1}{2}t^2(sin\frac{\pi}{4}) dt
    $

    = $\displaystyle \left[-\frac{1}{6} t^3 cos(\pi/4) \right]^1_0$ + $\displaystyle \left[-\frac{1}{6} t^3 sin(\pi/4) \right]^1_0$

    = $\displaystyle -\frac{1}{6}cos(\pi/4) + -\frac{1}{6}sin(\pi/4)$

    for r2(t)

    $\displaystyle
    \bold{r}_1(t) = t\cos \frac{\pi}{4}\bold{i}-t\sin \frac{\pi}{4}\bold{j} \mbox{ for }0\leq t\leq 1
    $

    i got: $\displaystyle \frac{1}{6}cos(\pi/4) + \frac{1}{6}sin(\pi/4)$


    SO THE LINE INTEGRALS OF r1(t) and r2(t) cancel each other out.

    for r3(t)... This is where im having problems

    $\displaystyle
    \bold{r}_3(t) = \cos t\bold{i} + \sin t \bold{j} \mbox{ for }-\frac{\pi}{4} \leq t\leq \frac{\pi}{4}
    $

    $\displaystyle \int_c xy dx + \int_c x^2 dy = \int^\frac{\pi}{4}_\frac{-\pi}{4}-cos(t)sin^2(t) dt
    $ + $\displaystyle \int^\frac{\pi}{4}_\frac{-\pi}{4}cos^3(t)
    $

    but i cant seem to do any more of it.. stuck here.. how do i complete it
    .................................................. .....
    ok the questions:
    (a) were my line integrals of r1 and r2 correct
    (b) what is the answer of r3
    (c) is the final answer of these line integrals $\displaystyle \frac{2}{3}sin(\pi/4)
    $
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  8. #8
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    Your question has been answered here.
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  9. #9
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    thnx mate i'll check it out now
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