Line Integral via Stokes' Theorem, Clockwise Orientation: 04-BS-1, May 2017, Q6

Question 6: 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. A closed curve $C$ (the tilted ellipse where the cylinder $x^2+y^2=1$ meets the plane $z=1+y$) and vector field $\mathbf v=(4z,-2x,2x)$, traversed clockwise as seen from above.

$C$ projects onto the unit circle $x^2+y^2=1$; the stated orientation is clockwise from above, opposite the standard (CCW/upward-normal) Stokes convention.

Find. $\displaystyle\oint_C\mathbf v\cdot d\mathbf r$ for the stated (clockwise) orientation.

Approach. Apply Stokes' theorem over the flat disk bounded by $C$ using the standard upward-normal (counterclockwise) orientation, then negate the result since the question specifies clockwise.

Curl of $\mathbf v$. $$\operatorname{curl}\mathbf v=\left(\frac{\partial(2x)}{\partial y}-\frac{\partial(-2x)}{\partial z},\ \frac{\partial(4z)}{\partial z}-\frac{\partial(2x)}{\partial x},\ \frac{\partial(-2x)}{\partial x}-\frac{\partial(4z)}{\partial y}\right)=(0,\,2,\,-2).$$
Surface and upward normal. Take $S$ as the flat disk $z=1+y$ over $x^2+y^2\le1$. Writing $z=g(x,y)=1+y$, the upward (CCW-consistent) normal element is $(-g_x,-g_y,1)=(0,-1,1)$, constant over the disk.
Dot with the curl and integrate (CCW value). $\operatorname{curl}\mathbf v\cdot(0,-1,1)=2(-1)+(-2)(1)=-4$, constant, so $$\oint_{C,\,\text{CCW}}\mathbf v\cdot d\mathbf r=\iint_D(-4)\,dA=-4\cdot\pi(1)^2=-4\pi.$$
Negate for the stated clockwise orientation. $$\boxed{\oint_{C,\,\text{CW}}\mathbf v\cdot d\mathbf r=4\pi}\approx12.57$$ (Check: direct parametrization $x=\cos\theta,y=\sin\theta,z=1+\sin\theta$, $\theta:0\to2\pi$ (CCW) gives $\oint\mathbf v\cdot d\mathbf r=\int_0^{2\pi}-4(1+\sin\theta)\sin\theta\,d\theta=-4\pi$, matching the Stokes value exactly.)

Quantity	Result
$\operatorname{curl}\mathbf v$	$(0,2,-2)$
CCW (standard) value	$-4\pi$
CW (as stated) value	$4\pi\approx12.57$

Concept & Theory

Topic. Stokes' theorem with a non-standard (clockwise) traversal orientation.

Key relations. $\displaystyle\oint_C\mathbf F\cdot d\mathbf r=\iint_S\operatorname{curl}\mathbf F\cdot d\mathbf S$ where $C$ is traversed counterclockwise relative to $S$'s chosen (upward) normal. Reversing $C$'s orientation negates the integral.

Why this works. The curl here is constant, so the flux integral collapses to (a constant) $\times$ (the disk's area) — no parametrized double integral is needed once the surface is chosen as the flat tilted disk rather than some more complicated cap.

Common pitfall. Computing the standard CCW value and reporting it directly without checking the question's stated orientation — this is the single most error-prone step in this problem type, always re-derive the flip/no-flip call rather than assuming a sign.

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

Question 6 of 8: Line Integral via Stokes' Theorem, Clockwise Orientation

Question 6: Line Integral via Stokes' Theorem, Clockwise Orientation 20 marks