Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stokes' Theorem - Limits of Integration

  1. Mar 26, 2007 #1
    Stokes' Theorem - Limits of Integration - Urgent! Please give a hand :)

    1. The problem statement, all variables and given/known data

    Assume the vector function

    [tex]\vec{A} = \hat{a}_x \left( 3x^2 y^3 \right) + \hat{a}_y \left( -x^3 y^2 \right)[/tex]

    Evaluate

    [tex] \int \left( \nabla \times \vec{A} \right) \cdot d\vec{s} [/tex]

    over the triangular area (see figure attached).

    2. Relevant equations

    Above.

    3. The attempt at a solution

    Step 1.

    [tex]\nabla \times \vec{A} & = \begin{vmatrix} \displaystyle \hat{a}_x & \displaystyle \hat{a}_y & \displaystyle \hat{a}_z \vspace{0.25cm} \\ \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \vspace{0.25cm} \\ \displaystyle A_x & \displaystyle A_y & \displaystyle 0 \end{vmatrix} \\
    & = \begin{vmatrix} \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\ \displaystyle A_y & \displaystyle 0 \end{vmatrix} \hat{a}_x - \begin{vmatrix} \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial z} \\ \displaystyle A_x & \displaystyle 0 \end{vmatrix} \hat{a}_y + \begin{vmatrix} \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} \\ \displaystyle A_x & \displaystyle A_y \end{vmatrix} \hat{a}_z \\
    & = \cancelto{0}{-\frac{\partial A_y}{\partial z} \, \hat{a}_x} + \cancelto{0}{\frac{\partial A_x}{\partial z} \, \hat{a}_y} + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \, \hat{a}_z[/tex]

    Step 2

    [tex]\frac{\partial A_y}{\partial x} = \frac{\partial}{\partial x} \left( -x^3 y^2 \right) = -y^2 \frac{d}{dx} \left( x^3 \right) = -3x^2 y^2[/tex]

    Step 3
    [tex]- \frac{\partial A_x}{\partial y} = - \frac{\partial}{\partial y} \left( 3x^2 y^3 \right) = -3x^2 \frac{d}{dy} \left( y^3 \right) = -9x^2y^2[/tex]

    Step 4
    [tex]\nabla \times \vec{A} & = \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) \, \hat{a}_z \\
    & = \left(-12x^2 y^2 \right) \, \hat{a}_z[/tex]

    Step 5

    [tex] \int \int \left(-12x^2 y^2 \right) \, \hat{a}_z \cdot \left( dx dy\right) \hat{a}_z [/tex]


    Not sure about the limits of integration - THIS IS "THE" PROBLEM. Maybe it's trivial and I just can't see it right now. I do expect to get one of the limits as a variable.

    I actually did the other way around (via Stokes' Theorem) P1 (1,1) to P2 (2,2) to P3 (2,1) to P1 and got

    [tex]\oint _S \vec{A} \cdot d\vec{\ell} = 7y^3 - 7x^3[/tex]

    I'm pretty sure this one is right, so I keep trying to make my other method work but without success so far. Any help is highly appreciated.

    THANKS, FOLKS!
     

    Attached Files:

    Last edited: Mar 27, 2007
  2. jcsd
  3. Mar 27, 2007 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Surely, you learned how to handle this long before you got to Stoke's theorem! The region is a triangle with vertices at (1,1), (2,1), and (2,2). If you choose to integrate with respect to y first, so that the outer integral is with respect to x, then x must cover the entire figure: x must go from 1 to 2. Now, for each x, y must go from the lower edge, y= 1 to the upper edge which is the line y= x. Your integral is
    [tex]\int_{x=1}^2 \int_{y=1}^x f(x,y)dydx[/tex]

    If, instead, you want to integrate with respect to x first so that the outer integral is with respect to y, then, again to cover the entire triangle, y must go from 1 to 2 and, for each y, x must go from the line x= y on the left to the line x= 2 on the right. Your integral is
    [tex]\int_{y= 1}^2 \int_{x=y}^2 f(x,y)dxdy[/tex]

     
  4. Mar 27, 2007 #3
    HallsofIvy, thanks for your comments.

    I actually tried those limits before, which sound reasonable to me - by the way. The problem is that I do get

    [tex] \int _1 ^2 \int _y ^2 \left(-12x^2 y^2 \right) dx dy = -\frac{98}{3} \neq \oint _S \vec{A} \cdot d\vec{\ell} = 7y^3 - 7x^3 [/tex]

    Hence, one of the sides is not correct. I double checked the work on the circulation and I can't see a mistake. Here's what I've got:

    Path of integration:

    [tex]P_1 (1,1) \to P_2 (2,2) \to P_3 (2,1) \to P_1 (1,1)[/tex]

    Step 1:

    [tex]\oint _S \vec{A} \cdot d\vec{\ell} & = \int _{P_1 P_2} \vec{A} \cdot d\vec{\ell} + \int _{P_2 P_3} \vec{A} \cdot d\vec{\ell} + \int _{P_3 P_4} \vec{A} \cdot d\vec{\ell}[/tex]


    Step 2:

    [tex]\int _{P_1 P_2} \vec{A} \cdot d\vec{\ell} & = \int _1 ^2 A_x \: dx + \int _1 ^2 A_y \: dy \\
    & = \int _1 ^2 \left( 3x^2 y^3 \right) \, dx + \int _1 ^2 \left( -x^3 y^2 \right) \, dy \\
    & = 3y^3 \int _1 ^2 x^2 \: dx - x^3 \int _1 ^2 y^2 \: dy \\
    & = 3y^3 \left. \frac{x^3}{3} \right| _1 ^2 - x^3 \left. \frac{y^3}{3} \right| _1 ^2 = 7y^3 - \frac{7}{3}x^3 \\[/tex]

    Step 3:

    [tex]\int _{P_2 P_3} \vec{A} \cdot d\vec{\ell} & = \int _2 ^2 A_x \: dx + \int _2 ^1 A_y \: dy \\
    & = -\int _1 ^2 \left( -x^3 y^2 \right) \, dy \\
    & = x^3 \int _1 ^2 y^2 \: dy \\
    & = x^3 \left. \frac{y^3}{3} \right| _1 ^2 = \frac{7}{3}x^3 \\[/tex]

    Step 4:

    [tex]\int _{P_3 P_2} \vec{A} \cdot d\vec{\ell} & = \int _2 ^1 A_x \: dx + \int _1 ^1 A_y \: dy \\
    & = -\int _1 ^2 \left( 3x^2 y^3 \right) \, dx \\
    & = -3y^3 \int _1 ^2 x^2 \: dx \\
    & = -3y^3 \left. \frac{x^3}{3} \right| _1 ^2 = - 7x^3\\[/tex]

    Step 5:

    [tex]\oint _S \vec{A} \cdot d\vec{\ell} = 7y^3 - 7x^3[/tex]

    Thanks again.
     
    Last edited: Mar 27, 2007
  5. Mar 27, 2007 #4

    HallsofIvy

    User Avatar
    Science Advisor

    You said you could do the path integral around the boundary! You do understand that the path integral should be a number, not a function of x and y?
    :rolleyes:
    I get 98/3 both ways- integrating over the surface and integrating around the boundary. (NOT -98/3: Since the path is clockwise around the surface, the normal is [itex]-\hat{a}_z[/itex], not [itex]\hat{a}_z[/itex]: "right hand rule".)

    I divided the boundary into three parts:
    I: The line from (2,1) to (1,1). Take x= t with t going from 2 to 1, y= 1. Then dx= dt, dy= 0 so the integral becomes
    [tex]\int_2^1 3t^2 dt= t^3 \right|_2^1= 1- 8= -7[/itex]

    II: The line from (2,2) to (2,1). Take x= 2, y= t with t going from 2 to 1. Then dx= 0, dy= dt so the integral becomes
    [tex]-8\int_2^1 t^2 dt= -\frac{8}{3}t^2\right|_2^1= \frac{8}{3}(1- 8)= -\frac{56}{3}[/tex]

    III: The line from (1,1) to (2,2). Take x= t, y= t with t going from 1 to 2. Then dx= dy= dt so the integral becomes
    [tex]\int_1^2 (3t^5- t^5)dt= 2\int_1^2 t^5 dt= \frac{1}{3}t^6\right|_1^2= \frac{1}{3} (64- 1)= \frac{63}{3}[/tex]

    So the complete integral is
    [tex]- 7+ \frac{56}{3}+ \frac{63}{3}= \frac{-21+ 56+ 63}{3}= \frac{98}{3}[/tex]
     
    Last edited by a moderator: Mar 27, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook