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hold on
i was just finished with line integrals =D

use Green's theorem...

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

I can try

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

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

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

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

hold on...

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

can someone help me with my problem

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

you with me?

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

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

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

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

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

then t would be between 0 and 2pi..

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

then we can do the change of variables

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

that's what we did

the problem is the double integral

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

what would be the bounds of integration then?

ok..i see

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

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

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

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

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

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

yes.

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

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

this is not going to work, is it?

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

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

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

sin squared

darn, you caused me soooo much hell.

this is much easier than i thought

I mean no harm. I come in peace.

jk =D

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

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

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

hmmm...

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

yes.

much better

lets start over

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

follow?

did i get the derivative wrong?

hold on

perfect
works out very nicely

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

ok

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

follow?

yesm

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

ok..

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

that s what got us into this mess ;)

now, you can do a change of variables!

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

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

64pi

im pretty sure

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

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

thats what i meant 2pi i still get 32pi

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

oooooh....ok

you are oh so smart

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

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