A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
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
 5 years ago
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

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on i was just finished with line integrals =D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use Green's theorem...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...wait, if and when im done with this one, can you help me w/ a parametrization?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so green's theorem says: doubleintegral(dQ/dxdP/dy)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, dQ/dx=4x*cos(2x)+2sin(2x) dP/dy=4sin(2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doubleintegral(4x*cos(2x)2sin(2x))dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh, wait you need to use a "power reducer"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(2x)=12sin^2(x) sin(2x)=2sin(x)cos(x) ...ugly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can someone help me with my problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doubleintegral(4x8x*sin^2(x)4sin(x)cos(x))dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i got you. But shouldnt we paramatize the x and y values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, that's not going to work... hold on

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0I was thinking that b/c this was a circle, we'll do a change of variables into polar coordinates

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so..x=4cos(t) y=4sin(t) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, but right now, as far as I am aware, you can't integrate cos(cos(t))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then t would be between 0 and 2pi..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have to work it out somehow so that the sines and cosines can disappear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we can do the change of variables

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so right now we are at doubleintegral(4x*cos(2x)2sin(2x))dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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
 5 years ago
Best ResponseYou've already chosen the best response.0the problem is the double integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0seems that i made a mistake when i did dQ/dx it's supposed to be 2cos(2x)+2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what would be the bounds of integration then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right now, we haven't "parametrized" in polar coordinates, the integral would be from 0 to 2pi, 0 to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's actually doubleintegral(4x*cos(2x)+2sin(2x)+2)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean... doubleintegral(2cos(2x)+4sin(2x)+2)dA oops, can't keep track

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0know of any property that can reduce this so that the trig disappears?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm..well is there an identity for cos(2x) or sin(2x) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, though i don't think it's going to help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you might try a classic integration, with y from sqrt(4x^2) to sqrt(4x^2) and x from 4 to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in that case you need a calculator to help you with arcsin and what not

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im doing that now, but the calc's going terribly slowly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is not going to work, is it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know I got a 32sqrt(4x^2) using wolfram mathmatics

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see, 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
 5 years ago
Best ResponseYou've already chosen the best response.0remember, it's a double integral, x should be eliminated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait a minute, that sin2x you wrote, it that a "sine squared x" or a "sine double x"???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0darn, you caused me soooo much hell.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is much easier than i thought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean no harm. I come in peace.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you rewrite the question a bit neater so i know what exactly im integrating?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"sine squared x" should be (sin(x))^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok..(4y)(sin(x))^2 i + (2x + sin 2x) j

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 2x + sin2x i presume is "sine double x"?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dQ/dx=2cos(2x)+2 dP/dy=4(sin(x))^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did i get the derivative wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0perfect works out very nicely

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so doubleintegral(dQ/dxdP/dy)dA is doubleintegral(2cos(2x)+2+4(sin(x))^2)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using cos(2x)=2(cos(x))^21 we get doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which is doubleintegral(4(cos(x))^22+2+4(sin(x))^2)dA which is doubleintegral(4(cos(x))^2+4(sin(x))^2)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, 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
 5 years ago
Best ResponseYou've already chosen the best response.0arrgh, i mean doubleintegral(2((2cos(x))^21)+2+4(sin(x))^2)dA

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you understand, though? didn't mean to have all those typos in the way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that s what got us into this mess ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0now, you can do a change of variables!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doubleintegral[theta,0,2pi][r,0,4](4r)dr dtheta evaluate that, and you're done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you get 32pi when you integrate r from 0 to 4 and theta from 0 to to pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, no r integrated is 16 pi, the full solution i worked out to be 64 pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you were supposed to integrate r from 0 to 4 and theta from 0 to 2pi, not pi.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i meant 2pi i still get 32pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integrate r, which is r^2/2. 4^2/20=8 integrate 1, which is theta 2pi0=2pi multiply you get 16 pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nahhh, it was nothing i got problems of my own.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'll post my question (for the 4th time) =D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.