Line Integral via Stokes' Theorem, Clockwise Orientation: 04-BS-1, May 2014, Q7

Question 7: Line Integral via Stokes' Theorem, Clockwise Orientation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Closed curve $C$: intersection of the cylinder $x^2+y^2=9$ (radius 3) with the plane $z=1+y-2x$, traversed clockwise viewed from $+z$. Vector field $\mathbf v=(4z,\,-2y,\,2y)$.

Find. $\displaystyle\oint_C\mathbf v\cdot d\mathbf r$.

$C$ = ellipse cut from the cylinder $x^2+y^2=9$ by the tilted plane $z=1+y-2x$, traced clockwise viewed from $+z$.

Approach. Apply Stokes' theorem over the flat elliptical patch of the plane bounded by $C$, projected onto the disk $x^2+y^2\le9$. Compute the curl, dot with the upward-oriented (CCW-paired) surface element, and negate at the end for the requested clockwise orientation.

Curl of $\mathbf v$. With $\mathbf v=(4z,\,-2y,\,2y)$, $$\nabla\times\mathbf v=\left(\frac{\partial(2y)}{\partial y}-\frac{\partial(-2y)}{\partial z},\ \frac{\partial(4z)}{\partial z}-\frac{\partial(2y)}{\partial x},\ \frac{\partial(-2y)}{\partial x}-\frac{\partial(4z)}{\partial y}\right)=(2,\ 4,\ 0).$$ Note this curl is a constant vector (every component of $\mathbf v$ is linear).
Upward-oriented surface element. For $z=f(x,y)=1+y-2x$, $d\mathbf S=(-f_x,-f_y,1)\,dx\,dy=(2,-1,1)\,dx\,dy$ (paired with counterclockwise-from-above by the right-hand rule).
Dot product — a constant integrand. $$(\nabla\times\mathbf v)\cdot(2,-1,1)=2(2)+4(-1)+0(1)=4-4=0.$$ Since both the curl and the surface-element direction are constant vectors, this dot product is identically zero everywhere on the patch — not merely zero after integrating an odd term by symmetry.
Integrate (trivial) and account for orientation. $\displaystyle\iint_{\text{disk}}0\,dA=0$ for the CCW-from-$+z$ orientation, so the requested clockwise integral is also $-0=0$.

$$\oint_C\mathbf v\cdot d\mathbf r=\boxed{0}$$

Quantity	Result
$\nabla\times\mathbf v$	$(2,4,0)$ (constant)
Upward surface element direction	$(2,-1,1)$
$\oint_C\mathbf v\cdot d\mathbf r$ (either orientation)	$0$

Concept & Theory

Topic. Stokes' theorem with a constant curl, and orientation bookkeeping.

Key relations. $\displaystyle\oint_{\partial S}\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot d\mathbf S$; for a linear vector field $\mathbf v$, $\nabla\times\mathbf v$ is a constant vector, so if that constant happens to be orthogonal to the (also constant, for a plane) surface-element direction, the flux integral — and hence the circulation — vanishes identically, independent of the patch's size or shape.

Why this works. Both the curl of a linear field and the normal of a plane are position-independent, so their dot product is a single number rather than a function of $(x,y)$; here that number happens to be exactly $0$, making the area of the ellipse irrelevant to the final answer.

Common pitfall. Assuming a zero result must be an error and second-guessing the curl computation; also, more generally, forgetting to negate a computed CCW-from-above value when the problem specifies clockwise (here it would not have mattered since the value is $0$, but the negation step must still be shown for full marks).

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 10 (§10.9 Stokes' Theorem).

Question 7 of 8: Line Integral via Stokes' Theorem, Clockwise Orientation

Question 7: Line Integral via Stokes' Theorem, Clockwise Orientation (20 marks)