Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane): 04-BS-1, May 2018, Q6

Question 6: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane) (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 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 $x^2+y^2\leq9$. Compute the curl once, then dot it against the (constant) surface-element vector.

Curl of $\mathbf v$. With $\mathbf v=(4z,\,-2y,\,2y)$, $$\nabla\times\mathbf v=\left(\frac{\partial(2y)}{\partial y}-\frac{\partial(-2y)}{\partial z},\ \frac{\partial(4z)}{\partial z}-\frac{\partial(2y)}{\partial x},\ \frac{\partial(-2y)}{\partial x}-\frac{\partial(4z)}{\partial y}\right)=(2,\,4,\,0).$$
Upward-oriented surface element. For $z=f(x,y)=1+y-2x$, the upward-normal surface element (paired with CCW-from-above by the right-hand rule) is $d\mathbf S=(-f_x,-f_y,1)\,dx\,dy=(2,-1,1)\,dx\,dy$.
Dot the curl into the surface element. $$(\nabla\times\mathbf v)\cdot(2,-1,1)=2(2)+4(-1)+0(1)=4-4+0=0.$$ The integrand vanishes identically over the entire disk, regardless of area or orientation.
Conclude. Since $(\nabla\times\mathbf v)\cdot d\mathbf S\equiv0$ pointwise, the surface integral — and hence the circulation, for either orientation — is exactly zero. The requested clockwise-from-above value equals the CCW value equals $0$; there is nothing to negate.

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

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

Concept & Theory

Topic. Stokes' theorem, including the case where curl and normal are orthogonal.

Key relations. $\displaystyle\oint_{\partial S}\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot d\mathbf S$. If $\nabla\times\mathbf v$ is orthogonal to the surface normal everywhere on a flat patch, the integrand is identically zero and the orientation choice becomes moot.

Why this works. Capping $C$ with the flat planar patch of $z=1+y-2x$ reduces the problem to a single dot product between the (constant, since $\mathbf v$ is linear) curl vector $(2,4,0)$ and the (also constant, for a planar patch) surface-element direction $(2,-1,1)$; these two vectors happen to be exactly perpendicular, so no integration is even needed once the dot product is taken.

Common pitfall. Assuming a zero boxed answer signals an arithmetic mistake and reworking the problem unnecessarily — a zero flux is a legitimate result whenever the curl is orthogonal to the surface normal, and should be reported as-is once independently verified. Also: forgetting that a zero circulation holds for both orientations, so the "assume clockwise" instruction in the problem statement changes nothing here.

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

Question 6 of 8: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane)

Question 6: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane) (20 marks)