Stoke's Theorem Validation with v = xy x + 2yz y + 3xz z, JDGriffiths Prob 1.33

[Living_Dog]

JDGriffiths, 3rd ed, Prob. 1.33:

Given v = xy x + 2yz y + 3xz z, check that Stoke's theorem is valid using the surface bounded by:
x = 0, 0 <= y <= 2, and 0 <= z = -y + 2. (See attached image.)

DJGriffiths Prob 1.33.bmp

Area Integral Result:
$$\int \nabla \times \mathbf{v} \cdot d\mathbf{a} = \int (-2y) (\frac{1}{2} dy dz) = -4$$

Line Integral Result:
$$\int \mathbf{v} \cdot d\mathbf{l} } = \int (xy dx + 2yz dy + 3 xz dz) = \frac{8}{3}$$

Stoke's theorem says they should be equal. What am I doing worng?

 

[xman]

check your integral over the surface, note that
$$d\vec{a} = dy \, dz \hat{x} \Rightarrow \vec{\nabla} \wedge \vec{v}\cdot d\vec{a} = 2 \,y\,dy\,dz$$
and watch the limits of the integral for dz. you should obtain the same results as your line integral. hope this helps. sincerely, x

 

[HallsofIvy]

JDGriffiths, 3rd ed, Prob. 1.33:

Given v = xy x + 2yz y + 3xz z, check that Stoke's theorem is valid using the surface bounded by:
x = 0, 0 <= y <= 2, and 0 <= z = -y + 2. (See attached image.)

DJGriffiths Prob 1.33.bmp

Area Integral Result:
$$\int \nabla \times \mathbf{v} \cdot d\mathbf{a} = \int (-2y) (\frac{1}{2} dy dz) = -4$$

Why "(1/2)dydz"? That area is in the yz-plane so the differential of area is just dydz.

Line Integral Result:
$$\int \mathbf{v} \cdot d\mathbf{l} } = \int (xy dx + 2yz dy + 3 xz dz) = \frac{8}{3}$$

Be careful of the direction in which you are integrating. I presume you see that it is only the line z= 2- y that gives a non-zero integral.

 

[Living_Dog]

check your integral over the surface, note that
$$d\vec{a} = dy \, dz \hat{x} \Rightarrow \vec{\nabla} \wedge \vec{v}\cdot d\vec{a} = 2 \,y\,dy\,dz$$
and watch the limits of the integral for dz. you should obtain the same results as your line integral. hope this helps. sincerely, x

I checked and get the same result. So here is what I do in detail:

$$\vec{\nabla} \times \vec{v} = \left| \begin{array}{ccc}<br /> \vec{x} & \vec{y} & \vec{z} \\<br /> \partial_{x} & \partial_{y} & \partial_{z} \\<br /> xy & 2yz & 3xz \\<br /> \end{array} \right| = (-2y) \hat{x} + ...$$

I still get a "-2y" for the curl since only the negative x component of the curl is non-zero and only that component is needed for the dot product with da.

I think I see my mistake now that you both have mentioned it - my area integral is messed up. I have to review area integrals and their limits.

Thanks for the help!
-LD

EDIT: I got it! I wasn't setting my limits correctly at all. I read Schaum's Outline Series on Advanced Calculus (Chapter 9, pg. 180) and solved it in seconds.

$$\int_{y=0}^{y=2} \int_{z=0}^{z=-y+2} 2y dydz = \frac{8}{3}$$ :!)

thx again!
-LD

 

[Living_Dog]

check your integral over the surface, note that
$$d\vec{a} = dy \, dz \hat{x} \Rightarrow \vec{\nabla} \wedge \vec{v}\cdot d\vec{a} = 2 \,y\,dy\,dz$$
and watch the limits of the integral for dz. you should obtain the same results as your line integral. hope this helps. sincerely, x

Oh, I am getting a negative sign when I do the curl. How are you getting a plus sign??

$$(\vec{\nabla} \times \vec{v})_{x} = \left| \begin{array}{ccc}<br /> \vec{x} & \vec{y} & \vec{z} \\<br /> \partial_{x} & \partial_{y} & \partial_{z} \\<br /> xy & 2yz & 3xz \\<br /> \end{array} \right| = 0 - 2y = -2y$$

 

[xman]

Oh, I am getting a negative sign when I do the curl. How are you getting a plus sign??

$$(\vec{\nabla} \times \vec{v})_{x} = \left| \begin{array}{ccc}<br /> \vec{x} & \vec{y} & \vec{z} \\<br /> \partial_{x} & \partial_{y} & \partial_{z} \\<br /> xy & 2yz & 3xz \\<br /> \end{array} \right| = 0 - 2y = -2y$$

The minus sign dissappears from the dot product with the $$d\vec{a} = -yz \hat{x}$$ which corresponds to the direction we are transversing the loop for the line integral.

 

[xman]

Oh, I am getting a negative sign when I do the curl. How are you getting a plus sign??

$$(\vec{\nabla} \times \vec{v})_{x} = \left| \begin{array}{ccc}<br /> \vec{x} & \vec{y} & \vec{z} \\<br /> \partial_{x} & \partial_{y} & \partial_{z} \\<br /> xy & 2yz & 3xz \\<br /> \end{array} \right| = 0 - 2y = -2y$$

The minus sign dissappears from the dot product with the $$d\vec{a} = -dydz \hat{x}$$ which corresponds to the direction we are transversing the loop for the line integral.