# Using Stokes' Theorem to evaluate over a trace on a plane

I'm trying to solve this problem that was given for my homework assignment, but I cannot figure out how to actually finish the problem in a way that makes sense to me. I've gotten as far as finding $\text{curl }\textbf{F} = \langle1, 1, 1\rangle$ from the original integral $\oint_C z dx + x dy + y dz$. (I believe that F is $\langle z, x, y \rangle$)

However, the question says to evaluate that integral over $C$, which is the trace of the cylinder $x^2 + y^2 = 25$ on the plane $y + z = 10$.

My textbook says to rearrange the plane equation so that it's $z = 10 - y$.

I then use the formula from the book (such that $g(x, y) = z = 10 - y$, and $\textbf{F} = P\vec{i} + Q\vec{j} + R\vec{k}$.)

That formula is $\iint_D (-P \frac{\partial g}{\partial x} - Q \frac{\partial g}{\partial y} + R) dA = \iint_S \textbf{F} \cdot dS$

Does this mean that $P = z$ (Obtained from $\textbf{F} = \langle z, x, y \rangle$) or does it mean that $P = 1$ (Obtained from $\text{curl }\textbf{F}$)? I'm leaning towards that it is $P = z$, but I end up with $0$ after setting up and solving.

My work uses the $P = z$ version so my integral looks like

$\iint_D (x + y) dA$

I then perform a change-of-coordinate-system on it resulting in $\int_0^{2\pi} \int_0^5 r^2 (\sin \theta + \cos \theta) dr d\theta = 0$

Is my answer correct or have I made an error in solving it? If so, how would I go about solving this properly?

$\oint_C \textbf{F}\cdot d\textbf{r} = \iint_S \nabla\times\textbf{F}\cdot d\textbf{A}$

$\textbf{F}=(z,x,y)$, $\nabla\times\textbf{F}=(1,1,1)$

Define a surface function for the plane, $G(x,y,z) = y+z-10$, the normal vector $\textbf{n}= \nabla G = (0,1,1)$

and write the differential area element in $dxdy$

$d\textbf{A} = \frac{\textbf{n}}{|n_3|} dxdy$

The integral is then , $\oint_C \textbf{F}\cdot d\textbf{r} = \iint_{x^2+y^2\le25} \nabla\times\textbf{F}\cdot \textbf{n} dxdy = (1,1,1)\cdot(0,1,1)\iint_{x^2+y^2\le25} dxdy = 50\pi$

• How did you get $G(x, y, z)$, as well as how does $2\iint_{x^2 + y^2 \le 25} dx dy$ become $50\pi$? do you re-write it as $2 \int_0^{2\pi} \int_0^5 r dr d\theta$? – Rebecca Jun 2 '16 at 21:04
• Whenever you have a surface defined by $g(x,y,z)=const.$, then $G(x,y,z) = g(x,y,z)-const$. The integral $\iint_{x^2 + y^2 \le 25} dx dy$ is simply the area of the circle ${x^2 + y^2 \le 25}$ – mastrok Jun 3 '16 at 5:45