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aarnes

  • 3 years ago

Solve line integral (xy+x-y)dy + (xy+x+y)dx where the curve is a circle x^2+y^2-ax=0

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  1. aarnes
    • 3 years ago
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    \[\int\limits_{?}^{?} (xy+x-y)dy + (xy+x+y)dx \] Circle formula: \[x ^{2} + y ^{2} - ax = 0\]

  2. henkjan
    • 3 years ago
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    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
    • 3 years ago
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    Are you sure? I think you need to use polar coordinates.

  4. henkjan
    • 3 years ago
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    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
    • 3 years ago
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    Ok, I'll try it and see how it goes.

  6. aarnes
    • 3 years ago
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    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
    • 3 years ago
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    Here's my try to solve it using polar coordinates.

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  8. aarnes
    • 3 years ago
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    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
    • 3 years ago
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    That's a*cos(theta)...Typo.

  10. aarnes
    • 3 years ago
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    I should've mentioned it. I'm using Green's theorem.

  11. TuringTest
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    since it is closed , consider green theorem

  15. TuringTest
    • 3 years ago
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    that's what we used, look at the scan above

  16. imranmeah91
    • 3 years ago
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    (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
    • 3 years ago
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    Nx=y+1

  18. imranmeah91
    • 3 years ago
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    \[\int \int (y-x ) dx dy\]

  19. imranmeah91
    • 3 years ago
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    just integrate over circle

  20. imranmeah91
    • 3 years ago
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    \[\int_0^{2\pi} \int_0^{ \frac{a}{2} cos(\theta)} a/2(sin(\theta))-(cos(\theta)) r dr d\theta\]

  21. TuringTest
    • 3 years ago
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    ^^^ 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
    • 3 years ago
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    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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