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Really simple line integral

  1. Aug 26, 2006 #1
    In my emag course we are reviewing vector calculus. I've forgotton a lot over the summer, so I just want to make sure I'm doing this properly.

    question)
    [tex] \vec E = \hat x y + \hat y x [/tex]
    Evaluate [itex] \int \vec E \cdot d\vec l [/itex] from [itex] P_1(2,1,-1) [/itex] to [itex] P_2(8,2,-1) [/itex] along the parabola [itex] x = 2y^2 [/itex].

    sol)
    We are in cartesian coordinates, thus:
    [tex] d\vec l = \hat x dx + \hat y dy [/tex]
    [tex] \vec E \cdot d\vec l = ydx + xdy [/tex]

    Our path is:
    [tex] x=2y^2 [/tex]
    [tex] y=\sqrt{\frac{x}{2}}[/tex]

    Thus,
    [tex] \int_2^8 \sqrt{\frac{x}{2}}\,\,dx + \int_1^2 2y^2 \,\,dy = \frac{28}{3}+\frac{14}{3}=14 [/tex]


    Does everything look ok?
     
    Last edited: Aug 26, 2006
  2. jcsd
  3. Aug 26, 2006 #2

    quasar987

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    To me yes.
     
  4. Aug 26, 2006 #3
    Cool. Another question if you wouldn't mind checking :)

    question)
    Evaluate,
    [tex] \oint_S \hat R 3 \sin \theta \cdot d\vec s [/tex]

    over the surface of a sphere with radius 5 centered at the orgin.

    ans)
    This question is in spherical coordinates. The notation used is in the format [itex] (R, \phi, \theta) [/itex]

    [tex]d\vec s = \hat R R^2 \sin \theta \,d\theta d\phi + \hat \phi R \,dR d\theta + \hat \theta R \sin \theta \,dR d\phi [/tex]

    [tex] \hat R3\sin \theta \cdot d\vec s = (R^2 \sin \theta d\theta d\phi)(3\sin \theta) = 3R^2 \sin^2 \theta d\theta d\phi [/tex]

    For our surface we have,
    [tex]R=5[/tex]
    [tex] 0 \leq \phi \leq 2\pi [/tex]
    [tex] 0 \leq \theta \leq \pi [/tex]

    The integral becomes,
    [tex] 3(25) \int_0^{2\pi} \int_0^\pi \sin^2 \theta \,\, d \theta d\phi = 75\pi^2 [/tex]

    Does this look ok also?
     
    Last edited: Aug 26, 2006
  5. Aug 26, 2006 #4

    quasar987

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    Correct also. Note that a faster approach is to see that since [itex]d\vec{s}[/itex] is the vector ds times the unit vector perpendicular to the surface, it is actually just

    [tex]\hat{R}ds=\hat{R}R^2sin\theta d\theta d\phi[/tex]
     
  6. Aug 26, 2006 #5
    Thanks man :smile:

    I always appreciate your help :)
     
  7. Aug 26, 2006 #6

    quasar987

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    No prob! It helps me to keep this stuff fresh in my memory too ;)
     
  8. Aug 27, 2006 #7

    HallsofIvy

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    Yes, but I wouldn't do it that way, because I dislike square roots!
    I would do everything in terms of y: [itex]\hat l= (2y^2)\hat x+ y\hat y[/itex] so [itex]d\hat l= (4y)dy \hat x+ dy \hat y[/itex] and [itex]\vec E= y \hat x+ x\hat y= y\hat x+ 2y^2 \hat y[/itex]

    [tex]\vec E \cdot d\hat l= (4y^2)dy+ (2y^2)dy= 6y^2 dy[/tex]
    Since, to go from (2, 1, -1) to (8, 2, -1), y must go from 1 to 2, the integral is
    [tex]\int_1^2 6y^2 dy= 2y^3\right|_1^2= 16- 2= 14[/itex]
    just as you got.
     
  9. Aug 27, 2006 #8
    Nice. I'll try this on some of the other review questions. Thank you :smile:
     
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