- #1
smoothman
- 39
- 0
how can greens theorem be verified for the region R defined by [tex](x^2 + y^2 \leq 1), (x + y \geq 0), (x - y \geq 0) ...
P(x,y) = xy, Q(x,y) = x^2 [/tex]
> okay i know [tex]\int_C Pdx + Qdy = \int\int \left(\frac{dQ}{dx} - \frac{dp}{dy}\right) dA[/tex]
so: [tex]\int_C xy dx + x^2dy = \int\int_D \left(2x - x\right) dy dx[/tex]
.........
here's my problem:
the limits of integration for the region D expressed as polar co-ordinates are:
[tex]-\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \mbox{ and }0\leq r\leq 1[/tex]
i understand "r" is from 0 to 1 because of the radius...
but how do you explain why theta is from -pi/4 to pi/4
-------------------------------
anyways using those limits for integration this is what i got:
ok so using polar co-ordinates:
2x - x = x
in polar terms: x = rcos\theta
so the integral is now:
[tex]\int_C xy dx + x^2dy = \int^{\pi/4}_{-\pi/4}\int^1_0 \left(rcos\theta\right)(r) drd\theta = \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}r^3cos\theta\right]^1_0d\theta = \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}cos\theta\right]d\theta = \left[\frac{1}{3}sin\theta\right]^{\pi/4}_{-\pi/4} = \frac{2}{3}sin(\pi/4)
[/tex]
is my result correct?
--------------------------
lastly, the curl integral can be shown using three line integrals:
[tex]
\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
[/tex]
[tex]
\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
[/tex]
[tex]
\bold{r}_3(t) = \cos t\bold{i} + \sin t \bold{j} \mbox{ for }-\frac{\pi}{4} \leq t\leq \frac{\pi}{4}
[/tex]
can you please explain why the line integrals are as they are.
P(x,y) = xy, Q(x,y) = x^2 [/tex]
> okay i know [tex]\int_C Pdx + Qdy = \int\int \left(\frac{dQ}{dx} - \frac{dp}{dy}\right) dA[/tex]
so: [tex]\int_C xy dx + x^2dy = \int\int_D \left(2x - x\right) dy dx[/tex]
.........
here's my problem:
the limits of integration for the region D expressed as polar co-ordinates are:
[tex]-\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \mbox{ and }0\leq r\leq 1[/tex]
i understand "r" is from 0 to 1 because of the radius...
but how do you explain why theta is from -pi/4 to pi/4
-------------------------------
anyways using those limits for integration this is what i got:
ok so using polar co-ordinates:
2x - x = x
in polar terms: x = rcos\theta
so the integral is now:
[tex]\int_C xy dx + x^2dy = \int^{\pi/4}_{-\pi/4}\int^1_0 \left(rcos\theta\right)(r) drd\theta = \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}r^3cos\theta\right]^1_0d\theta = \int^{\pi/4}_{-\pi/4}\left[\frac{1}{3}cos\theta\right]d\theta = \left[\frac{1}{3}sin\theta\right]^{\pi/4}_{-\pi/4} = \frac{2}{3}sin(\pi/4)
[/tex]
is my result correct?
--------------------------
lastly, the curl integral can be shown using three line integrals:
[tex]
\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
[/tex]
[tex]
\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
[/tex]
[tex]
\bold{r}_3(t) = \cos t\bold{i} + \sin t \bold{j} \mbox{ for }-\frac{\pi}{4} \leq t\leq \frac{\pi}{4}
[/tex]
can you please explain why the line integrals are as they are.