anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hold on i was just finished with line integrals =D
anonymous
  • anonymous
use Green's theorem...
anonymous
  • anonymous
...wait, if and when im done with this one, can you help me w/ a parametrization?

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anonymous
  • anonymous
I can try
anonymous
  • anonymous
so green's theorem says: doubleintegral(dQ/dx-dP/dy)dA
anonymous
  • anonymous
so, dQ/dx=4x*cos(2x)+2sin(2x) dP/dy=-4sin(2x)
anonymous
  • anonymous
doubleintegral(4x*cos(2x)-2sin(2x))dA
anonymous
  • anonymous
ahh, wait you need to use a "power reducer"
anonymous
  • anonymous
hold on...
anonymous
  • anonymous
cos(2x)=1-2sin^2(x) sin(2x)=2sin(x)cos(x) ...ugly
anonymous
  • anonymous
can someone help me with my problem
anonymous
  • anonymous
doubleintegral(4x-8x*sin^2(x)-4sin(x)cos(x))dA
anonymous
  • anonymous
you with me?
anonymous
  • anonymous
yeah i got you. But shouldnt we paramatize the x and y values
anonymous
  • anonymous
wait, that's not going to work... hold on
anonymous
  • 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
  • anonymous
I was thinking that b/c this was a circle, we'll do a change of variables into polar coordinates
anonymous
  • anonymous
so..x=4cos(t) y=4sin(t) ?
anonymous
  • anonymous
yeah, but right now, as far as I am aware, you can't integrate cos(cos(t))
anonymous
  • anonymous
then t would be between 0 and 2pi..
anonymous
  • anonymous
we have to work it out somehow so that the sines and cosines can disappear
anonymous
  • anonymous
then we can do the change of variables
anonymous
  • anonymous
so right now we are at doubleintegral(4x*cos(2x)-2sin(2x))dA
anonymous
  • 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
  • anonymous
that's what we did
anonymous
  • anonymous
the problem is the double integral
anonymous
  • anonymous
seems that i made a mistake when i did dQ/dx it's supposed to be 2cos(2x)+2
anonymous
  • anonymous
what would be the bounds of integration then?
anonymous
  • anonymous
right now, we haven't "parametrized" in polar coordinates, the integral would be from 0 to 2pi, 0 to 4
anonymous
  • anonymous
ok..i see
anonymous
  • anonymous
it's actually doubleintegral(4x*cos(2x)+2sin(2x)+2)dA
anonymous
  • anonymous
i mean... doubleintegral(2cos(2x)+4sin(2x)+2)dA oops, can't keep track
anonymous
  • anonymous
know of any property that can reduce this so that the trig disappears?
anonymous
  • anonymous
hmm..well is there an identity for cos(2x) or sin(2x) ?
anonymous
  • anonymous
yes, though i don't think it's going to help.
anonymous
  • 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
  • anonymous
yes.
anonymous
  • anonymous
in that case you need a calculator to help you with arcsin and what not
anonymous
  • anonymous
im doing that now, but the calc's going terribly slowly
anonymous
  • anonymous
this is not going to work, is it?
anonymous
  • anonymous
i dont know I got a 32sqrt(4-x^2) using wolfram mathmatics
anonymous
  • 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
  • anonymous
remember, it's a double integral, x should be eliminated
anonymous
  • anonymous
wait a minute, that sin2x you wrote, it that a "sine squared x" or a "sine double x"???
anonymous
  • anonymous
sin squared
anonymous
  • anonymous
darn, you caused me soooo much hell.
anonymous
  • anonymous
this is much easier than i thought
anonymous
  • anonymous
I mean no harm. I come in peace.
anonymous
  • anonymous
jk =D
anonymous
  • anonymous
can you rewrite the question a bit neater so i know what exactly im integrating?
anonymous
  • anonymous
"sine squared x" should be (sin(x))^2
anonymous
  • anonymous
ok..(-4y)(sin(x))^2 i + (2x + sin 2x) j
anonymous
  • anonymous
hmmm...
anonymous
  • anonymous
and 2x + sin2x i presume is "sine double x"?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
much better
anonymous
  • anonymous
lets start over
anonymous
  • anonymous
dQ/dx=2cos(2x)+2 dP/dy=-4(sin(x))^2
anonymous
  • anonymous
follow?
anonymous
  • anonymous
did i get the derivative wrong?
anonymous
  • anonymous
hold on
anonymous
  • anonymous
perfect works out very nicely
anonymous
  • anonymous
so doubleintegral(dQ/dx-dP/dy)dA is doubleintegral(2cos(2x)+2+4(sin(x))^2)dA
anonymous
  • anonymous
ok
anonymous
  • anonymous
using cos(2x)=2(cos(x))^2-1 we get doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA
anonymous
  • 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
  • anonymous
follow?
anonymous
  • anonymous
yesm
anonymous
  • 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
  • anonymous
arrgh, i mean doubleintegral(2((2cos(x))^2-1)+2+4(sin(x))^2)dA
anonymous
  • anonymous
ok..
anonymous
  • anonymous
you understand, though? didn't mean to have all those typos in the way.
anonymous
  • anonymous
that s what got us into this mess ;)
anonymous
  • 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
  • anonymous
now, you can do a change of variables!
anonymous
  • anonymous
doubleintegral[theta,0,2pi][r,0,4](4r)dr dtheta evaluate that, and you're done
anonymous
  • anonymous
do you get 32pi when you integrate r from 0 to 4 and theta from 0 to to pi
anonymous
  • anonymous
64pi
anonymous
  • anonymous
im pretty sure
anonymous
  • anonymous
well, no r integrated is 16 pi, the full solution i worked out to be 64 pi
anonymous
  • anonymous
you were supposed to integrate r from 0 to 4 and theta from 0 to 2pi, not pi.
anonymous
  • anonymous
thats what i meant 2pi i still get 32pi
anonymous
  • 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
  • anonymous
oooooh....ok
anonymous
  • anonymous
you are oh so smart
anonymous
  • anonymous
nahhh, it was nothing i got problems of my own.
anonymous
  • anonymous
i'll post my question (for the 4th time) =D

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