Question

Use green's theorem to evaluate the line integral along the given positive orientated curve.
\[\int_c xe^{-2x}dx+(x^4+2x^2y^2)dy\]
C is the boundary of the region between the circles \[x^2+y^2=1\] and \[x^2+y^2=4\]
\[\int\int_D\left[\frac{\partial}{\partial x}(x^4+2x^2y^2)-\frac{\partial}{\partial y}(xe^{-2x})\right]dA \]
\[\int\int_D(4x^3+4xy^2)dA\]
would I do greens theorem again?
\[\int\int_D\left[\frac{\partial}{\partial x}(4xy^2)-\frac{\partial}{\partial y}(4x^3)\right]dA \]
\[\int\int_D4y^2dA\]

Answer 1

to my understanding for last stage u just need to compute double integral

Answer 2

yeah, and I would do it in polar coordinates

Answer 3

\[\iint\limits_D(4x^3+4xy^2)dA\]\[D:\{(r,\theta)|1\le r\le2,0\le\theta\le2\pi\}\]\[x=r\cos\theta~~~~~~~y=r\sin\theta~~~~~~~~dA=rdrd\theta\]

Answer 4

I understand everything you wrote except for \[dA=rdrd\theta\]

Answer 5

that is dA in polar coordinates...
if you want to prove that you need to use the jacobian (not sure if you're doing that yet)

Answer 6

not yet

Answer 7

so then you have to take my word for it that you need to add another r in the dA for polar coordinates
or http://en.wikipedia.org/wiki/Polar_coordinate_system

Answer 8

fair enough.
So now we have:
\[\int_1^2\int_0^{2\pi}(4x^3+4xy^2)rdrd\theta\]

Answer 9

Nope that's wrong, let me try it again

Answer 10

gotta sub in for x and y

Answer 11

\[\int_1^2\int_0^{2\pi}(4(rcos\theta)^3+4(rcos\theta)(rsin\theta)^2)rdrd\theta\]

Answer 12

\[\int_1^2\int_0^{2\pi}(4r^4cos^3+4r^4cos\theta sin^2\theta)drd\theta\]

Answer 13

yep
now use some identities

Answer 14

well, one identity
just factor it right...

Answer 15

would I take the 4r^4 out of the first integral since we're doing the integral with respect to theta first?

Answer 16

try taking out 4r^4cos(theta)

Answer 17

I mean factoring it that way rather

Answer 18

that would give me \[cos^2\theta+sin^2\theta=1\] and now i would take the integral of
\[\int_1^2\int_0^{2\pi}4r^4cos\theta drd\theta\]

Answer 19

yes, and you will see quickly what it becomes

Answer 20

we would be left with \[\int_1^24r^4dr\] because the integral of cosine is sine and then (0-0)
\[4\left[\frac{r^5}{5}\right]_1^2=4(\frac{32}{5}-\frac15)\]

Answer 21

what is\[\int_0^{2\pi}\cos\theta d\theta\]?

Answer 22

is sine

Answer 23

and after evaluating...?

Answer 24

sine of 2pi is zero and so is sine of zero

Answer 25

and so that's where we end, because the integral of zero is zero
so the next one goes away....

Answer 26

oh that would make the double integral zero?

Answer 27

yep

Answer 28

I hate when that happens; always makes me think I messed up
but here I don't think so

Answer 29

oh ok. Thanks! Does that mean there is no area between the region bounded by those two circles?

Answer 30

no, it means that when you integrate that function over the area the answer is zero.

obviously the area between a circle radius 1 and radius 2 itself is not zero; it can easily be calculated with old-fashioned geometry

Answer 31

which, according to green's theorem, means that the line integral is zero

Answer 32

I thought we were integrating to find the area?

Answer 33

oh sorry we're evaluating the curve

Answer 34

no, and area integral would be\[\iint\limits_D dA\]and nothing more
that is different than\[\iint\limits_Df(x)dA\]in which case you are integrating a \(function\) over an area

Answer 35

greens theorem relates a line integral over a closed curve to a surface integral of a related function

Answer 36

\[\oint\limits_C Pdx+Qdy=\iint\limits_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA\]

Answer 37

it's a relation between the two ideas; a closed line integral and the surface integral on the right.

Answer 38

I think I'm starting to get the idea now. I'll do more problems to get a better grip on greens theorem.

Answer 39

here's a good reference:
http://tutorial.math.lamar.edu/Classes/CalcIII/GreensTheorem.aspx

Answer 40

I can't believe I forgot to reference Paul again!

Answer 41

yeah, apul has a similar problem in example 3
a note about LaTeX:
to make the double integral look better, and to get the D underneath the signs, try
\iint\limits_D\[\iint\limits_D\]
likewise
\iiint\[\iiint\] is a triple integral
and
\oint\[\oint\]is a line integral

Answer 42

paul*

Answer 43

Yeah i thought my integrals were quite far apart. I'll do that for this next problem i'm working on. Thanks buddy!

Answer 44

\oint is a *closed* line integral I mean..

Answer 45

anytime!

Answer 46

dude I forgot to give you a medal!