Question

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

Answer 1

hold on
i was just finished with line integrals =D

Answer 2

use Green's theorem...

Answer 3

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

Answer 4

I can try

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

hold on...

Answer 10

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

Answer 11

can someone help me with my problem

Answer 12

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

Answer 13

you with me?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

then t would be between 0 and 2pi..

Answer 21

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

Answer 22

then we can do the change of variables

Answer 23

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

Answer 24

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

Answer 25

that's what we did

Answer 26

the problem is the double integral

Answer 27

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

Answer 28

what would be the bounds of integration then?

Answer 29

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

Answer 30

ok..i see

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

yes.

Answer 38

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

Answer 39

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

Answer 40

this is not going to work, is it?

Answer 41

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

Answer 42

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.

Answer 43

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

Answer 44

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

Answer 45

sin squared

Answer 46

darn, you caused me soooo much hell.

Answer 47

this is much easier than i thought

Answer 48

I mean no harm. I come in peace.

Answer 49

jk =D

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

hmmm...

Answer 54

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

Answer 55

yes.

Answer 56

much better

Answer 57

lets start over

Answer 58

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

Answer 59

follow?

Answer 60

did i get the derivative wrong?

Answer 61

hold on

Answer 62

perfect
works out very nicely

Answer 63

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

Answer 64

ok

Answer 65

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

Answer 66

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

Answer 67

follow?

Answer 68

yesm

Answer 69

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

Answer 70

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

Answer 71

ok..

Answer 72

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

Answer 73

that s what got us into this mess ;)

Answer 74

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

Answer 75

now, you can do a change of variables!

Answer 76

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

Answer 77

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

Answer 78

64pi

Answer 79

im pretty sure

Answer 80

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

Answer 81

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

Answer 82

thats what i meant 2pi i still get 32pi

Answer 83

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

Answer 84

oooooh....ok

Answer 85

you are oh so smart

Answer 86

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

Answer 87

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