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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)
ahh, wait you need to use a "power reducer"
cos(2x)=1-2sin^2(x) sin(2x)=2sin(x)cos(x) ...ugly
can someone help me with my problem
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 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
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
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
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?
right now, we haven't "parametrized" in polar coordinates, the integral would be from 0 to 2pi, 0 to 4
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
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
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.
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"???
darn, you caused me soooo much hell.
this is much easier than i thought
I mean no harm. I come in peace.
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
and 2x + sin2x i presume is "sine double x"?
lets start over
did i get the derivative wrong?
perfect works out very nicely
so doubleintegral(dQ/dx-dP/dy)dA is doubleintegral(2cos(2x)+2+4(sin(x))^2)dA
using cos(2x)=2(cos(x))^2-1 we get doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA
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
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
arrgh, i mean doubleintegral(2((2cos(x))^2-1)+2+4(sin(x))^2)dA
you understand, though? didn't mean to have all those typos in the way.
that s what got us into this mess ;)
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
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
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
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