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anonymous

  • 4 years ago

calculus question below involving green's theorem.

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{C}^{}(x^2-xy)dx+(xy-y^2)dy\] clockwise around the triangle with the vertices (0,0), (1,1), (2,0)

  2. anonymous
    • 4 years ago
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    |dw:1328595663698:dw|

  3. anonymous
    • 4 years ago
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    essentially I only need to know how to set up the limits of integration

  4. anonymous
    • 4 years ago
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    Well what are you trying to do, exactly? Evaluate this as a line integral or a surface integral?

  5. anonymous
    • 4 years ago
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    i think it's a surface, I'm not exactly sure though, that is all the question asks.

  6. anonymous
    • 4 years ago
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    what's the general way to differentiate between the two?

  7. anonymous
    • 4 years ago
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    does a circle in the middle of the integration sign mean anything?

  8. anonymous
    • 4 years ago
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    |dw:1328596021608:dw|

  9. anonymous
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    When you perform an integral, you're taking a function that's defined over some domain, breaking the domain up into little chunks, and adding it all together. When the domain is one-dimensional, then we call it a line integral. The standard integral you learned when you first started calculus is just a specific example of a line integral where the line happens to be the x- axis, but you can generalize to any line you wish, including the boundary of that triangle that you drew.

  10. anonymous
    • 4 years ago
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    On the other hand, a surface integral is performed when your domain is two-dimensional. Such would be the case if you were integrating over the interior of the triangle, rather than its boundary: |dw:1328596318103:dw|

  11. anonymous
    • 4 years ago
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    Green's theorem is a very important result from Vector Calculus that is a special case of another, vastly more general theorem that actually governs all of the integral calculus you've ever done and all you're ever likely to do (though it won't be presented as such) known as Stokes' Theorem.

  12. anonymous
    • 4 years ago
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    The idea behind it is basically that if you have a function F, which happens to be a vector, then the following two integrals are equal: \[\oint \vec{F}\cdot d\vec{r} = \iint \vec{\nabla}\cdot \vec{F}\space dS\]

  13. anonymous
    • 4 years ago
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    oops, that should be \[\vec{\nabla} \times \vec{F} \], not the dot..

  14. anonymous
    • 4 years ago
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    this is actualy helping a lot, i have had a hard time truly grasping the concepts here.

  15. anonymous
    • 4 years ago
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    This may appear somewhat foreign to you at the moment, as you may be seeing new symbols and everything, but I'll simplify. If we let \[d\vec{r} = <dx,dy> \] and \[ \vec{F} = <P,Q> \] and the surface S simply be the xy plane, then the above equation becomes \[ \oint Pdx + Qdy = \iint \left( \frac{dQ}{dx} - \frac{dP}{dy}\right)dxdy\]

  16. anonymous
    • 4 years ago
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    This is more likely the form you've seen the equation in. It seems somewhat weird and arbitrary, but it's relatively easy to prove and see if you'd like me to show you. In words it says: The line integral of a function F around a closed loop of some kind is equal to the surface integral of the curl of the function evaluated over the surface enclosed by the loop.

  17. anonymous
    • 4 years ago
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    Let's take your specific example: for you, \[P = x^2 - xy\] and \[Q = xy - y^2\] Evaluating this as a line integral would be tedious and difficult. However, we can apply Green's theorem to make our lives easier. \[\frac{dP}{dy} = -x\] \[\frac{dQ}{dx} = y\] so our original integral is equal to, by Green's theorm, \[\iint \left(\frac{dQ}{dx} - \frac{dP}{dy}\right) dxdy =\iint \left(y + x\right)dxdy \]

  18. anonymous
    • 4 years ago
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    great! that helps so much in making it clearer, So how would I finish it off from here?

  19. anonymous
    • 4 years ago
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    Okay, now you just need to evaluate that double integral. You can evaluate the integral in either order( either x or y first) but one would be easier than the other. Notice that if you evaluate the y integral first, the bounds on your integrals would be [ y goes from the x axis to the line above it] [x goes from zero to two] The reason we would like to steer clear of this is because we would need to do TWO integrals. This is because "the line above it" is piecewisely defined, so we would have to do an integral for the left part, which is the line y = x, and another for the right part, which is the line y = 2 - x

  20. anonymous
    • 4 years ago
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    right okay.. that makes sense..

  21. anonymous
    • 4 years ago
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    so instead.. 0<=y<=1 y<=x<=1-y ........?

  22. anonymous
    • 4 years ago
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    i feel like that might probably be wrong.. aha.

  23. anonymous
    • 4 years ago
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    very close :) should be 2-y, not 1-y

  24. anonymous
    • 4 years ago
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    So let's see this baby in action. With our limits plugged in, that should be \[ \int_0^1 \int_{y}^{2-y}(x + y) dxdy\]

  25. anonymous
    • 4 years ago
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    Evaluating the first integral yields \[\int_0^1 \left[ \frac{x^2}{2} + xy\right] |^{2-y}_y dy = \int_0^1 \left[\frac{(2-y)^2}{2} - y(2-y)\right]-\left[\frac{y^2}{2} - y^2 \right] dy \] \[= \int_0^1 2y^2 -3y +4 \space dy = \left[\frac{2y^3}{3} - \frac{3y^2}{2} + 4y\right]^1_0 = 2/3 - 3/2 +4 = 19/6\] unless I made a typo :) Which I'm sure I almost inevitably did, but that's the idea.

  26. anonymous
    • 4 years ago
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    aha thank you SO much, for your help, Your method helped me to arrive at the correct answer. i got more from you in 20 minutes then reading out of this difficult Adam's calculus book for hours!

  27. anonymous
    • 4 years ago
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    I'm glad I could be of help :) If you'd ever like to see why Green's Theorem is true (you might even like it, it's delightfully simple and kind of cute) let me know. And good luck with the rest of your work, we'll be around if you need us!

  28. anonymous
    • 4 years ago
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    hey that's great! you know I'll be looking for you with questions ! perhaps one more quick one that is just technical jumble, what's the exact method of converting to spherical coordinates like what does x = y= z= and what variables are added to the integrand in terms of r, phi and theta

  29. anonymous
    • 4 years ago
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    if that makes sense..

  30. anonymous
    • 4 years ago
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    it seems like half the questions i do this year, deal in spherical or cylindrical, and i wanna just understand them so i quite avoiding them.

  31. anonymous
    • 4 years ago
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    Since this is a math class, I assume you use the angle theta for the angle in the xy plane, and phi for the angle descending from the z axis, is that correct?

  32. anonymous
    • 4 years ago
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    yes

  33. anonymous
    • 4 years ago
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    And do you use rho as the distance from the origin? Some people use rho, other people use r...

  34. anonymous
    • 4 years ago
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    i think in our text book it has r but i've seen both notations

  35. anonymous
    • 4 years ago
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    alright well I'll use r just to be safe. \[ x = r\sin(\phi)\cos(\theta) \] \[ y = r\sin(\phi)\sin(\theta) \] \[z = r\cos(\phi) \]

  36. anonymous
    • 4 years ago
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    As far as converting your integrals, that added factor is called the Jacobian. Do you know why it exists?

  37. anonymous
    • 4 years ago
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    I can't say i wouldn't mind an explanation... ahah

  38. anonymous
    • 4 years ago
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    Alright, do me a favor, and start a new thread, will you?

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