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Need help with some vector calculus problems

  1. Jan 25, 2005 #1
    f = 2xy in the x dir, (x^2 - z^2) in the y dir, -3xz^2 in the z dir.
    particle travles from (0,0,0) to (2,1,3)

    im supposed to find the line integral from (0,0,0) to (2,1,3).
    if i make z = 3/2 x and y = 1/2 x and substitute those into the integral of F dot dl i end up with an answer of -27.66667. the way im doing it makes sense from the things ive seen on the net. but the book gives an answer of -39.5. ??

    also, ive got another problem, my electro-magnetics prof didnt get a chance to do any examples on this stuff. D = 2 Ro z^2 in the Ro dir, and Ro cos^2 theta in the z dir.
    find closed surface integral of D dot Ds.
    0<Ro<5
    -1<z<1
    0<theta<2pi

    ive got no idea on this one. is it basically the same thing as a line integral or do i do something different?
     
  2. jcsd
  3. Jan 26, 2005 #2

    HallsofIvy

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    I have no idea what you mean by "f = 2xy in the x dir, (x^2 - z^2) in the y dir, -3xz^2 in the z dir". Do you possibly mean 2xy dx+ (x2- z2)dy- 3xz2dz ? Oh, wait, you mean f is a vector: f= (2xy)i+ (x2- z2)j- 3xz2k.
    Of course its the same thing: fds= 2xy dx+ (x2- z2)dy- 3xz2dz
    To integrate it from (0, 0, 0) to (2, 1, 3), let x= 2t, y= t, z= 3t so that dx= 2dt,
    dy= dt, dz= 3dt and substitute. The integral will be for t= 0 to 1.

    I'm a little put off by using "Ro" instead of r in cylindrical coordinates. That's normally reserved for polar coordinates.

    the figure "0< Ro< 5, -1< z< 1, 0< theta< 2p is the cylinder with radius 5, extending from z= -1 to 1. It's surface (dS) really has three parts: the top, a disk of radius 5 with z= 1 (and dS= rdrd&theta;), the bottom, a disk of radius 5 with z= -1 (and dS= -rdr&theta;), and the "lateral side" with dS= rdrd&theta;.
     
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