anonymous
  • anonymous
Solve line integral (xy+x-y)dy + (xy+x+y)dx where the curve is a circle x^2+y^2-ax=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{?}^{?} (xy+x-y)dy + (xy+x+y)dx \] Circle formula: \[x ^{2} + y ^{2} - ax = 0\]
anonymous
  • anonymous
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)
anonymous
  • anonymous
Are you sure? I think you need to use polar coordinates.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
Ok, I'll try it and see how it goes.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Here's my try to solve it using polar coordinates.
1 Attachment
anonymous
  • anonymous
I switched from ax as limit to a+cos(theta) during integration. And also I used 2pi instead of -(pi/2) to (pi/2)
anonymous
  • anonymous
That's a*cos(theta)...Typo.
anonymous
  • anonymous
I should've mentioned it. I'm using Green's theorem.
TuringTest
  • TuringTest
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?
TuringTest
  • TuringTest
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.
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
since it is closed , consider green theorem
TuringTest
  • TuringTest
that's what we used, look at the scan above
anonymous
  • anonymous
(xy+x-y)dy + (xy+x+y)dx N dy + M dx Nx= y-1 My= x+1 \[\int \int Nx- My dx dy\]
TuringTest
  • TuringTest
Nx=y+1
anonymous
  • anonymous
\[\int \int (y-x ) dx dy\]
anonymous
  • anonymous
just integrate over circle
anonymous
  • anonymous
\[\int_0^{2\pi} \int_0^{ \frac{a}{2} cos(\theta)} a/2(sin(\theta))-(cos(\theta)) r dr d\theta\]
TuringTest
  • TuringTest
^^^ 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?
anonymous
  • anonymous
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!

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