# Line Integral and Greens Theorem

• Nov 30th 2011, 09:00 AM
dbakeg00
Line Integral and Greens Theorem
I'm supposed to do the following problems, first using line integrals and then Green's theorem....but I keep getting different answers. Can someone please help me see my error? Thanks.

1a) Let C be the positively oriented boundry of the first quadrant part of the unit disk. Use the definition of the line integral to find:

$\displaystyle \int (xy)dx + (x+y)dy$

My Work:
$\displaystyle x=cos(t), y=sin(t), 0\leq t \leq \frac{\pi}{2}$ b/c it is only in the first quadrant

$\displaystyle \vec{F}<x,y>=(cos(t)*sin(t))$

$\displaystyle \vec{G}<x,y>=(cos(t)+sin(t))$

$\displaystyle \int_0^\frac{\pi}{2}[(cos(t)*sin(t))(-sin(t))+(cos(t)+sin(t))(cos(t))]dt$

$\displaystyle =\int_0^\frac{\pi}{2}[-sin^2(t)*cos(t)+cos^2(t)+sin(t)*cos(t)]dt$

$\displaystyle =\int[(-(u)^2)+u)]du+\int_0^\frac{\pi}{2}[\frac{1}{2}+\frac{cos(2t)}{2}]dt$

$\displaystyle [-\frac{sin^3(t)}{3}+\frac{sin^2(t)}{2}+\frac{t}{2}+ \frac{sin(2t)}{4}]\right|_0^\frac{}\pi{2}$

$\displaystyle =[-\frac{1}{3}+\frac{1}{2}+\frac{\pi}{4}]$

$\displaystyle =\frac{\pi}{4}-\frac{1}{6}$

1b) Now using Green's Theorem:

$\displaystyle \frac{\partial G}{\partial x}=1$

$\displaystyle \frac{\partial F}{\partial y}=x$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[1-r cos(\theta)]r*dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[r-r^2 cos(\theta)]dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{r^2}{2}-\frac{r^3}{3}cos(\theta)]\right|_0^1 d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{1}{2}-\frac{1}{3}cos(\theta)]d\theta$

$\displaystyle =[\frac{1}{2}\theta-\frac{1}{3}sin(\theta)]\right|_0^\frac{\pi}{2}$

$\displaystyle =\frac{\pi}{4}-\frac{1}{3}$
• Nov 30th 2011, 09:16 AM
HallsofIvy
Re: Line Integral and Greens Theorem
Quote:

Originally Posted by dbakeg00
I'm supposed to do the following problems, first using line integrals and then Green's theorem....but I keep getting different answers. Can someone please help me see my error? Thanks.

1a) Let C be the positively oriented boundry of the first quadrant part of the unit disk. Use the definition of the line integral to find:

$\displaystyle \int (xy)dx + (x+y)dy$

My Work:
$\displaystyle x=cos(t), y=sin(t), 0\leq t \leq \frac{\pi}{2}$ b/c it is only in the first quadrant

$\displaystyle \vec{F}<x,y>=(cos(t)*sin(t))$

$\displaystyle \vec{G}<x,y>=(cos(t)+sin(t))$

$\displaystyle \int_0^\frac{\pi}{2}[(cos(t)*sin(t))(-sin(t))+(cos(t)+sin(t))(cos(t))]dt$

$\displaystyle =\int_0^\frac{\pi}{2}[-sin^2(t)*cos(t)+cos^2(t)+sin(t)*cos(t)]dt$

$\displaystyle =\int[(-(u)^2)+u)]du+\int_0^\frac{\pi}{2}[\frac{1}{2}+\frac{cos(2t)}{2}]dt$

$\displaystyle [-\frac{sin^3(t)}{3}+\frac{sin^2(t)}{2}+\frac{t}{2}+ \frac{sin(2t)}{4}]\right|_0^\frac{}\pi{2}$

$\displaystyle =[-\frac{1}{3}+\frac{1}{2}+\frac{\pi}{4}]$

$\displaystyle =\frac{\pi}{4}-\frac{1}{6}$

$\displaystyle \frac{1}{2}- \frac{1}{3}= +\frac{1}{6}$ not $\displaystyle -\frac{1}{6}$.

Quote:

1b) Now using Green's Theorem:

$\displaystyle \frac{\partial G}{\partial x}=1$

$\displaystyle \frac{\partial F}{\partial y}=x$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[1-r cos(\theta)]r*dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[r-r^2 cos(\theta)]dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{r^2}{2}-\frac{r^3}{3}cos(\theta)]\right|_0^1 d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{1}{2}-\frac{1}{3}cos(\theta)]d\theta$

$\displaystyle =[\frac{1}{2}\theta-\frac{1}{3}sin(\theta)]\right|_0^\frac{\pi}{2}$

$\displaystyle =\frac{\pi}{4}-\frac{1}{3}$
..
Green's theorem says that area integral is equal to the integral of your first function over the entire boundary of the region. That includes the integral from (0, 0) to (1, 0) along the x-axis and the integral from (0, 1) to (0, 0) along the y-axis.
• Nov 30th 2011, 09:40 AM
dbakeg00
Re: Line Integral and Greens Theorem
So are you saying my bounds are wrong? I see that I have a typo on the double integrals bounds, they should have read 0 to 1.What should my bounds be?
• Dec 1st 2011, 04:20 AM
HallsofIvy
Re: Line Integral and Greens Theorem
No, I am saying that your path integral is not around the entire boundary of the region!
• Dec 1st 2011, 05:12 AM
dbakeg00
Re: Line Integral and Greens Theorem
I now see what you meant...I only covered the arc portion in my first attempt. When I included the axis portions, I ended up getting the same asnwer as I did using Green's Theorem. Thanks for the help.