Line Integral via Stokes' Theorem, Clockwise Orientation: 04-BS-1, December 2014, Q5

Question 5: 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 $x^2+y^2=1$ (radius 1) with the plane $z=1+y$, traversed clockwise viewed from $+z$. Vector field $\mathbf v=(4z,\,-2x,\,2x)$.

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

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

Approach. Apply Stokes' theorem over the flat elliptical patch of the plane bounded by $C$, projected onto the unit disk $x^2+y^2\le1$. 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 $\mathbf v=(4z,\,-2x,\,2x)$, $$\nabla\times\mathbf v=\left(\frac{\partial(2x)}{\partial y}-\frac{\partial(-2x)}{\partial z},\ \frac{\partial(4z)}{\partial z}-\frac{\partial(2x)}{\partial x},\ \frac{\partial(-2x)}{\partial x}-\frac{\partial(4z)}{\partial y}\right)=(0,\ 2,\ -2).$$
Upward-oriented surface element. For $z=f(x,y)=1+y$, $d\mathbf S=(-f_x,-f_y,1)\,dx\,dy=(0,-1,1)\,dx\,dy$ (paired with counterclockwise-from-above by the right-hand rule).
Dot product — a constant integrand. $$(\nabla\times\mathbf v)\cdot(0,-1,1)=0(0)+2(-1)+(-2)(1)=-4.$$ Both the curl and the surface-element direction are constant vectors, so this dot product is $-4$ everywhere on the patch.
Integrate over the unit disk (CCW-from-above), then flip for the requested orientation. $$\iint_{\text{disk}}(-4)\,dA=-4\cdot\pi(1)^2=-4\pi\quad\text{(CCW from }+z\text{)}.$$ The problem specifies the clockwise orientation, the reverse of CCW, so negate: $$\oint_C\mathbf v\cdot d\mathbf r=-(-4\pi)=4\pi.$$

$$\oint_C\mathbf v\cdot d\mathbf r=\boxed{4\pi}$$

Quantity	Result
$\nabla\times\mathbf v$	$(0,2,-2)$ (constant)
Upward surface element direction	$(0,-1,1)$
CCW-from-above value	$-4\pi$
Requested (CW-from-above) value	$4\pi$

Concept & Theory

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

Key relations. $\displaystyle\oint_{\partial S}\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot d\mathbf S$, with the surface orientation and boundary traversal linked by the right-hand rule; reversing the traversal direction negates the result.

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 constant number, reducing the surface integral to that constant times the patch's area — here the unit disk's area $\pi$.

Common pitfall. Forgetting to negate the CCW-from-above result when the problem specifies a clockwise traversal — this sign flip is the single most error-prone step in this problem type and must always be shown explicitly, never silently absorbed.

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

Question 5 of 8: Line Integral via Stokes' Theorem, Clockwise Orientation

Question 5: Line Integral via Stokes' Theorem, Clockwise Orientation (20 marks)