I was doing some homework problems on using Stokes' Theorem and I got the correct answer for all but one of the problems. I looked around at other questions related to Stokes' Theorem on here and figured out that I needed to divide by the magnitude of the normal vector, and that indeed brought me to the right answer, but I don't understand why I don't require that step for any of the other problems. I found explanations for why the magnitude of the normal vector cancels out, but I can't figure out why it didn't in this problem. I'll start with the problem I was having trouble with and then include one of the problems I solved correctly for context.
A circle C in the plane x+y+z=1 has a radius of 8 and center (2, -2, 1). Evaluate $\oint_C\mathbf F\cdot d\mathbf r$ for $\mathbf F = \langle 0, -4z, -2y\rangle$, where C has counterclockwise orientation when viewed from above.
Here's my attempted solution:
Start by finding the curl of F: $$ \mathbf F = \langle 0, -4z, -2y\rangle ; \space \nabla\times\mathbf F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & -4z & -2y \\ \end{vmatrix} =\langle2,0,0\rangle \\ $$ Parameterize the plane: $$ r(x,y)=\langle x,y,1-x-y\rangle \\ $$ Find the partial derivatives of r: $$ r_x(x,y)=\langle 1,0,-1\rangle \\ r_y(x,y)=\langle 0,1,-1\rangle \\ $$ Find the vector normal to r: $$ \mathbf n = r_x\times r_y = \begin{vmatrix} i & j & k \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ \end{vmatrix} = \langle 1,1,1 \rangle \text{ (vector is pointing in the correct direction)} $$ Calculate the integrand: $$ (\nabla \times \mathbf F)\cdot \mathbf n = 2(1)+0(1)+0(1) = 2$$ So our integral is: $$ \iint 2dxdy $$ Convert to polar coordinates so that we can set our bounds of integration as a circle: $$ \int_0^{2\pi}\int_0^8 2rdrd\theta$$ Solve the integral to get: $$ 128\pi $$ But according to the solution the correct answer is: $$ \frac{128\pi}{\sqrt{3}} $$
Here's a similar example where I used the same method to get the correct answer:
Confirm that Stokes’ Theorem holds for the vector field $F=\langle z-y, x, -x\rangle$, where S is the hemisphere $x^2+y^2+z^2=4$, for $z\ge 0$, and C is the circle $x^2+y^2=4$ oriented counterclockwise.
Start by finding the curl of F: $$ \mathbf F=\langle z-y,x,-x\rangle ;\space \nabla \times \mathbf F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z-y & x & -x \\ \end{vmatrix} = \langle 0,2,2 \rangle $$ Parameterize the surface: $$ r(x,y)=\langle x, y, \sqrt{4-x^2-y^2}$$ Find the partial derivatives of r: $$ r_x(x,y)= \langle 1,0, \frac{-x}{\sqrt{4-x^2-y^2}}\rangle \\ r_y(x,y)= \langle 0,1, \frac{-y}{\sqrt{4-x^2-y^2}}\rangle$$ Find the vector normal to r: $$ \mathbf n = r_x\times r_y = \begin{vmatrix} i & j & k \\ 1 & 0 & \frac{-x}{\sqrt{4-x^2-y^2}} \\ 0 & 1 & \frac{-y}{\sqrt{4-x^2-y^2}} \\ \end{vmatrix} = \langle \frac{x}{\sqrt{4-x^2-y^2}}, \frac{y}{\sqrt{4-x^2-y^2}}, 1\rangle$$ Calculate the integrand: $$ (\nabla \times \mathbf F)\cdot \mathbf n = 0(\frac{x}{\sqrt{4-x^2-y^2}}) + 2(\frac{y}{\sqrt{4-x^2-y^2}}) + 2(1) = \frac{2y}{\sqrt{4-x^2-y^2}} + 2$$ So our integral is: $$ \iint (\frac{2y}{\sqrt{4-x^2-y^2}} + 2) dxdy$$ Convert to polar coordinates so that we can set our bounds of integration as a circle: $$ \int_0^2 \int_0^{2\pi} (\frac{2rsin\theta}{\sqrt{4-r^2}} + 2)rd\theta dr = \int_0^2 \int_0^{2\pi} (\frac{2r^2sin\theta}{\sqrt{4-r^2}} + 2r)d\theta dr $$ In this case, integrating theta first makes the first term drop out and the integral reduces down to: $$ \int_0^2 4\pi rdr = 8\pi$$
In this case $8\pi$ was the correct answer, so I'm really unsure what I'm missing. I've been trying to figure this out basically all day today. Any help would be greatly appreciated.
