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Line integral of vector field from Apostol calculus

  1. Jun 3, 2017 #1
    1. The problem statement, all variables and given/known data

    Here are the three problems that i couldn't solve from the book Calculus volume 2 by apostol

    10.9 Exercise

    2. Find the amount of work done by the force f(x,y)=(x^2-y^2)i+2xyj in moving a particle (in a counter clockwise direction) once around the square bounded by the coordinate axes and the lines x=a and y=a, a>0

    4) A force field f in 3-space is give by the formula f(x,y,z)=yzi+xzj+x(y+1)k. Calculate the work done by f in moving a particle once around the triangle with verticles (0,0,0),(1,1,1),(-1,1,-1) in that order

    6) Calculate the work done by the force field f(x,y,z)=(y-z)i+(z-x)j+(x-y)k along the curve of intersection of the sphere x^2+y^2+z^2=4 and the plane z=ytanb where 0<b<pi/2. The path is tranversed in a direction that appears counterclockwise when viewed from high above th xy-plane.
    2. Relevant equations

    3. The attempt at a solution


    2)(0,0)-->(a,0) r(t)=ati r'(t)=ai
    (a,0)-->(a,a) r(t)=atj r'(t)=aj
    (a,a)-->(0,a) r(t)=-ati r'(t)=-ai
    (0,a)-->(0,0) r(t)=-atj r'(t)=-aj

    ∫ƒ.dr= ∫(a^2 t^2)i⋅ai dt (0-->1)+ ∫(-a^2t^2)i⋅aj dt + ∫(a^2t^2)i -ai⋅dt (-1-->0) + ∫(-a^2t^2)i⋅-aj dt
    =2a^3/3
    yet the correct answer is 2a^3 by the textbook

    4)(0,0,0)-->(1,1,1) r(t)=ti+tj+tk r'(t)=i+j+k
    (1,1,1)-->(-1,1,-1) r(t)=(1-2t)i+(1-2t)k r'(t)=-2i+-2k
    ∫ƒ.dr= ∫t^2i+t^2j+(t^2+t)k⋅(i+j+k) dt (0-->1) + ∫(1-2t)^2i +(1-2t)k ⋅(-2i-2k) dt (0-->1)
    =t^3+(5/2)t^2-2t
    =3/2
    yet the correct answer is 0 by the textbook

    6)i stuck in parametrize the equation
     
  2. jcsd
  3. Jun 3, 2017 #2

    Ray Vickson

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    (2) Using the simpler parametrizations ##t = x## or ##t = y## on the segments, we have:
    $$\begin{array}{l} W_{(0,0) \to (a,0)} = \int_0^a x^2 \, dx\\
    W_{(a,0) \to (a,a)} = \int_0^a 2 a y \, dy\\
    W_{(a,a) \to (0,a)} = \int_0^a (-) [(a-x)^2 - a^2] \, dx \\
    W_{(0,a) \to (0,0)} = 0
    \end{array}
    $$
    I get the book's value.

    (4) ##\vec{F } = \vec{F_1} + \vec{F_2}##, where ##\vec{F_1} = \langle yz, zx ,xy\rangle## and ##\vec{F_2} = \langle 0,0 ,x \rangle##.
    Note that ##\vec{F_1} = \vec{\nabla} V##, where ##V = xyz##, so the contribution of ##\vec{F_1}## to the total work is ##\int_C \vec{\nabla} V \cdot \,d \vec{r} = 0##, by Stokes theorem. Thus, we need only look at ##\int_C \langle 0,0, x \rangle \cdot \langle dx,dy,dz \rangle##, which is a lot easier than the original problem. Of course, this simplification does not help much if you have not yet seen Stokes theorem.

    (6) What type of ##xy## curve do you see if you view it from above the ##xy## plane? How would you parametrize such a curve?
     
    Last edited: Jun 3, 2017
  4. Jun 3, 2017 #3

    LCKurtz

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    Another hint for ##6##: If you calculate it you will find ##\nabla \times \vec f = \langle -2,-2,-2\rangle ##. You might try using Stokes' theorem that ##\int_C \vec f \cdot d\vec r =\iint_S \nabla \times \vec f \cdot d\vec S## where ##S## is the plane circular cross-section.
     
    Last edited: Jun 3, 2017
  5. Jun 3, 2017 #4

    vela

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    The parameterization ##r(t) = at \hat{j}## doesn't move you from (a,0) to (a,a). It goes from (0,0) to (0,a).
     
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