Stokes' Theorem - Limits of Integration

DivGradCurl · Mar 26, 2007

Stokes' Theorem - Limits of Integration - Urgent! Please give a hand :)

Homework Statement

Assume the vector function

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

Evaluate

\int \left( \nabla \times \vec{A} \right) \cdot d\vec{s}

over the triangular area (see figure attached).

Homework Equations

Above.

The Attempt at a Solution

Step 1.

\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

Step 2

\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

Step 3
- \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

Step 4
\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

Step 5

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

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

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!

HallsofIvy · Mar 27, 2007

thiago_j said:

Homework Statement

Assume the vector function

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

Evaluate

\int \left( \nabla \times \vec{A} \right) \cdot d\vec{s}

over the triangular area (see figure attached).

Homework Equations

Above.

The Attempt at a Solution

Step 1.

\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

Step 2

\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

Step 3
- \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

Step 4
\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

Step 5

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

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.

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
\int_{x=1}^2 \int_{y=1}^x f(x,y)dydx

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
\int_{y= 1}^2 \int_{x=y}^2 f(x,y)dxdy

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

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

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!

DivGradCurl · Mar 27, 2007

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

\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

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:

P_1 (1,1) \to P_2 (2,2) \to P_3 (2,1) \to P_1 (1,1)

Step 1:

\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}Step 2:

\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 \\

Step 3:

\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 \\

Step 4:

\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\\

Step 5:

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

Thanks again.

HallsofIvy · Mar 27, 2007

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?

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 -\hat{a}_z, not \hat{a}_z: "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
\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 -8\int_2^1 t^2 dt= -\frac{8}{3}t^2\right|_2^1= \frac{8}{3}(1- 8)= -\frac{56}{3} 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 \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} So the complete integral is - 7+ \frac{56}{3}+ \frac{63}{3}= \frac{-21+ 56+ 63}{3}= \frac{98}{3}

Stokes' Theorem - Limits of Integration

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Homework Statement

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