## aarnes Group Title Solve line integral (xy+x-y)dy + (xy+x+y)dx where the curve is a circle x^2+y^2-ax=0 2 years ago 2 years ago

1. aarnes Group Title

$\int\limits_{?}^{?} (xy+x-y)dy + (xy+x+y)dx$ Circle formula: $x ^{2} + y ^{2} - ax = 0$

2. henkjan Group Title

just integrate: integral of (xy+x-y)dy = 1/2*x*y^2 + xy - 1/2y^2 integral of (xy+x+y)dx = 1/2*x^2*y + 1/2x^2 + yx combining: 1/2*(x*y^2 - y^2+x^2*y + x^2) + 2*yx = 1/2*(-y^2+x^2) +2*x*y+1/2*(x^2*y+y^2*x)

3. aarnes Group Title

Are you sure? I think you need to use polar coordinates.

4. henkjan Group Title

You can always applly polar coordinates later on as well.... It might make it a little more cumbersome, but it will give the same value. But you can try, indeed.

5. aarnes Group Title

Ok, I'll try it and see how it goes.

6. aarnes Group Title

Well, I did same as you (nothing new) but I'm confused as to what limits to use. If converting to polar coordinates at start before integration I have one limit for angle theta 0 - 2pi and other for r 0 - a*cos(theta). With the result you got I don't know where to us these limits.

7. aarnes Group Title

Here's my try to solve it using polar coordinates.

8. aarnes Group Title

I switched from ax as limit to a+cos(theta) during integration. And also I used 2pi instead of -(pi/2) to (pi/2)

9. aarnes Group Title

That's a*cos(theta)...Typo.

10. aarnes Group Title

I should've mentioned it. I'm using Green's theorem.

11. TuringTest Group Title

Ok, leave the angle bounds as -pi/2 to pi/2. r varies from 0 to acos(theta). I got$\oint_c(xy+x-y)dy+(xy+x+y)dx=-\frac{a^3\pi}{8}$seemed rather straightforward, not sure about what the problem is. Is this not the answer?

12. TuringTest Group Title

If you wanted to change the bounds of theta to 0-2pi then you would at least have to make r vary from 0-a/2, but that still won't work as far as I can see. You would need to change more things in order to do that; seems extraneous.

13. TuringTest Group Title

btw you were doing that last integral on your paper strangely. you could have integrated cos^3x*sinx right away, then tackled cos^4x with a reduction formula.

14. imranmeah91 Group Title

since it is closed , consider green theorem

15. TuringTest Group Title

that's what we used, look at the scan above

16. imranmeah91 Group Title

(xy+x-y)dy + (xy+x+y)dx N dy + M dx Nx= y-1 My= x+1 $\int \int Nx- My dx dy$

17. TuringTest Group Title

Nx=y+1

18. imranmeah91 Group Title

$\int \int (y-x ) dx dy$

19. imranmeah91 Group Title

just integrate over circle

20. imranmeah91 Group Title

$\int_0^{2\pi} \int_0^{ \frac{a}{2} cos(\theta)} a/2(sin(\theta))-(cos(\theta)) r dr d\theta$

21. TuringTest Group Title

^^^ that could work too, but why do you have a/2 in the integrand? shouldn't it be r first? plus the circle you are integrating on imran is centered at (0,0), so won't that change things? I did$\oint_c(xy+x-y)dy+(xy+x+y)dx=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{a\cos\theta}r^2(\sin\theta-\cos\theta)drd\theta$and got my answer above$-\frac{a^3\pi}{8}$is there something wrong with that?

22. aarnes Group Title

Sorry for not answering for so long but I was away from keyboard. I too tried to do the problem using the r limit from 0 to a*cos(theta) and I got the same answer as TuringTest so I guess it's correct. Medal given. Thank you all for your input!