Does anyone know how to do this one?
Evaluate the line integral
∫ F dr
C
about the simple closed curve C, oriented in the positive direction, where C is the circle x2 + y2 = 16 and F = (-4y)(sin2 x) i + (2x + sin 2x) j

- anonymous

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- anonymous

hold on
i was just finished with line integrals =D

- anonymous

use Green's theorem...

- anonymous

...wait, if and when im done with this one, can you help me w/ a parametrization?

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## More answers

- anonymous

I can try

- anonymous

so green's theorem says:
doubleintegral(dQ/dx-dP/dy)dA

- anonymous

so, dQ/dx=4x*cos(2x)+2sin(2x)
dP/dy=-4sin(2x)

- anonymous

doubleintegral(4x*cos(2x)-2sin(2x))dA

- anonymous

ahh, wait
you need to use a "power reducer"

- anonymous

hold on...

- anonymous

cos(2x)=1-2sin^2(x)
sin(2x)=2sin(x)cos(x)
...ugly

- anonymous

can someone help me with my problem

- anonymous

doubleintegral(4x-8x*sin^2(x)-4sin(x)cos(x))dA

- anonymous

you with me?

- anonymous

yeah i got you. But shouldnt we paramatize the x and y values

- anonymous

wait, that's not going to work... hold on

- anonymous

i guess you know more about this than i do, we didn't parametrize anything when we were doing these. yes, i figure you can do that

- anonymous

I was thinking that b/c this was a circle, we'll do a change of variables into polar coordinates

- anonymous

so..x=4cos(t)
y=4sin(t) ?

- anonymous

yeah, but right now, as far as I am aware, you can't integrate cos(cos(t))

- anonymous

then t would be between 0 and 2pi..

- anonymous

we have to work it out somehow so that the sines and cosines can disappear

- anonymous

then we can do the change of variables

- anonymous

so right now we are at
doubleintegral(4x*cos(2x)-2sin(2x))dA

- anonymous

oh wait I see.. using greens theorem M or P=-4(sin^2(x))
N or Q=2x+sin2x
then, we derive with respect to x on N and derive with respect to y on M

- anonymous

that's what we did

- anonymous

the problem is the double integral

- anonymous

seems that i made a mistake when i did dQ/dx
it's supposed to be 2cos(2x)+2

- anonymous

what would be the bounds of integration then?

- anonymous

right now, we haven't "parametrized"
in polar coordinates, the integral would be from 0 to 2pi, 0 to 4

- anonymous

ok..i see

- anonymous

it's actually
doubleintegral(4x*cos(2x)+2sin(2x)+2)dA

- anonymous

i mean...
doubleintegral(2cos(2x)+4sin(2x)+2)dA
oops, can't keep track

- anonymous

know of any property that can reduce this so that the trig disappears?

- anonymous

hmm..well is there an identity for cos(2x) or sin(2x) ?

- anonymous

yes, though i don't think it's going to help.

- anonymous

you might try a classic integration, with y from -sqrt(4-x^2) to sqrt(4-x^2) and x from -4 to 4

- anonymous

yes.

- anonymous

in that case you need a calculator to help you with arcsin and what not

- anonymous

im doing that now, but the calc's going terribly slowly

- anonymous

this is not going to work, is it?

- anonymous

i dont know I got a 32sqrt(4-x^2) using wolfram mathmatics

- anonymous

see, now if we did the problem without using green's theorem, it won't work anyway, since F(r(t)) gets us another cosine within a cosine.

- anonymous

remember, it's a double integral, x should be eliminated

- anonymous

wait a minute, that sin2x you wrote, it that a "sine squared x" or a "sine double x"???

- anonymous

sin squared

- anonymous

darn, you caused me soooo much hell.

- anonymous

this is much easier than i thought

- anonymous

I mean no harm. I come in peace.

- anonymous

jk =D

- anonymous

can you rewrite the question a bit neater so i know what exactly im integrating?

- anonymous

"sine squared x" should be (sin(x))^2

- anonymous

ok..(-4y)(sin(x))^2 i + (2x + sin 2x) j

- anonymous

hmmm...

- anonymous

and 2x + sin2x i presume is "sine double x"?

- anonymous

yes.

- anonymous

much better

- anonymous

lets start over

- anonymous

dQ/dx=2cos(2x)+2
dP/dy=-4(sin(x))^2

- anonymous

follow?

- anonymous

did i get the derivative wrong?

- anonymous

hold on

- anonymous

perfect
works out very nicely

- anonymous

so doubleintegral(dQ/dx-dP/dy)dA
is doubleintegral(2cos(2x)+2+4(sin(x))^2)dA

- anonymous

ok

- anonymous

using cos(2x)=2(cos(x))^2-1
we get doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA

- anonymous

which is doubleintegral(4(cos(x))^2-2+2+4(sin(x))^2)dA
which is doubleintegral(4(cos(x))^2+4(sin(x))^2)dA

- anonymous

follow?

- anonymous

yesm

- anonymous

ok, oops, typo in
doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA:
should be doubleintegral(2(2cos(x)-1)^2+2+4(sin(x))^2)dA

- anonymous

arrgh, i mean
doubleintegral(2((2cos(x))^2-1)+2+4(sin(x))^2)dA

- anonymous

ok..

- anonymous

you understand, though?
didn't mean to have all those typos in the way.

- anonymous

that s what got us into this mess ;)

- anonymous

ok so now we have
doubleintegral(4(cos(x))^2+4(sin(x))^2)dA
easy cake:
(cos(x))^2+(sin(x))^2=1,
so, it is just doubleintegral(4)dA

- anonymous

now, you can do a change of variables!

- anonymous

doubleintegral[theta,0,2pi][r,0,4](4r)dr dtheta
evaluate that, and you're done

- anonymous

do you get 32pi when you integrate r from 0 to 4 and theta from 0 to to pi

- anonymous

64pi

- anonymous

im pretty sure

- anonymous

well, no r integrated is 16 pi,
the full solution i worked out to be 64 pi

- anonymous

you were supposed to integrate r from 0 to 4 and theta from 0 to 2pi, not pi.

- anonymous

thats what i meant 2pi i still get 32pi

- anonymous

integrate r, which is r^2/2.
4^2/2-0=8
integrate 1, which is theta
2pi-0=2pi
multiply you get 16 pi

- anonymous

oooooh....ok

- anonymous

you are oh so smart

- anonymous

nahhh, it was nothing
i got problems of my own.

- anonymous

i'll post my question
(for the 4th time) =D

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