Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane): 04-BS-1, December 2017, Q5

Question 5: 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 $z=5-x$, traversed clockwise viewed from $+z$. Vector field $\mathbf v=(xy,\,2z,\,3y)$.

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

$C$ = ellipse cut from the cylinder $x^2+y^2=9$ by the plane $z=5-x$ (tilted only in the $x$-$z$ direction), 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 account for orientation: compute the CCW-from-above (upward-normal) result first, then negate it because the problem specifies the opposite (clockwise) traversal.

Curl of $\mathbf v$. With $\mathbf v=(xy,\,2z,\,3y)$, $$\nabla\times\mathbf v=\left(\frac{\partial(3y)}{\partial y}-\frac{\partial(2z)}{\partial z},\ \frac{\partial(xy)}{\partial z}-\frac{\partial(3y)}{\partial x},\ \frac{\partial(2z)}{\partial x}-\frac{\partial(xy)}{\partial y}\right)=(3-2,\ 0,\ -x)=(1,\,0,\,-x).$$
Upward-oriented surface element. For $z=f(x,y)=5-x$, the upward-normal surface element (paired with CCW-from-above by the right-hand rule) is $d\mathbf S=(-f_x,-f_y,1)\,dx\,dy=(1,0,1)\,dx\,dy$.
Dot and integrate over the disk (CCW case first). $$(\nabla\times\mathbf v)\cdot(1,0,1)=1(1)+0(0)+(-x)(1)=1-x.$$ Over the disk $x^2+y^2\leq9$ (polar, $r\in[0,3]$), the $-x$ term integrates to zero by symmetry, leaving just the area $\pi(3)^2=9\pi$: $$\oint_{C,\,\text{CCW from }+z}\mathbf v\cdot d\mathbf r=9\pi.$$
Flip for the requested (clockwise) orientation. The problem specifies $C$ traversed clockwise viewed from $+z$, the opposite of the CCW result above, so the answer is the negative: $$\oint_C\mathbf v\cdot d\mathbf r=-9\pi.$$

$$\oint_C\mathbf v\cdot d\mathbf r=\boxed{-9\pi}$$

Quantity	Result
$\nabla\times\mathbf v$	$(1,0,-x)$
CCW-from-$+z$ integral	$9\pi$
Requested (CW) integral	$-9\pi$

Concept & Theory

Topic. Stokes' theorem, with careful orientation bookkeeping.

Key relations. $\displaystyle\oint_{\partial S}\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot d\mathbf S$, where the surface normal and the boundary traversal direction are linked by the right-hand rule.

Why this works. Capping $C$ with the flat planar patch of the plane $x+z=5$ (rather than some curved surface) reduces the surface integral to a simple double integral over the projected disk $x^2+y^2\le9$. Because the plane has no $y$-dependence, the $x$-part of the integrand is odd about $x=0$ over the symmetric disk and vanishes, leaving only the constant term times the area — and because the requested traversal is clockwise (the reverse of the "upward normal" pairing), the computed value must be negated.

Common pitfall. Forgetting to negate for the clockwise orientation, or negating when the curve was actually specified counterclockwise. Also sign errors extracting $d\mathbf S=(-f_x,-f_y,1)\,dA$ from $z=f(x,y)$.

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

Question 5 of 8: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane)

Question 5: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane) (20 marks)