Green's Theorem

• Aug 7th 2007, 04:08 PM
smoothman
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?
• Aug 7th 2007, 05:25 PM
ThePerfectHacker
Look at picture below.

This can be expressed in polar coordinates as:
$\displaystyle -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \mbox{ and }0\leq r\leq 1$
• Aug 7th 2007, 06:30 PM
joanne_q
Quote:

Originally Posted by ThePerfectHacker
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.
• Aug 7th 2007, 06:39 PM
smoothman
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)
• Aug 7th 2007, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by smoothman
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.

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

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

$\displaystyle \bold{r}_3(t) = \cos t\bold{i} + \sin t \bold{j} \mbox{ for }-\frac{\pi}{4} \leq t\leq \frac{\pi}{4}$
• Aug 8th 2007, 05:37 AM
smoothman
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?
• Aug 8th 2007, 12:59 PM
smoothman
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 $\displaystyle \frac{2}{3}sin(\pi/4)$
• Aug 8th 2007, 02:00 PM
ThePerfectHacker