Line Integral via Stokes' Theorem: 04-BS-1, December 2013, Q6

Question 6: Line Integral via Stokes' Theorem (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 $x^2+y^2=4$ (radius 2) with the plane $z=1+2x-y$, traversed clockwise viewed from $+z$. Vector field $\mathbf v=(z+x,\ -2y,\ y^2)$.

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

$C$ = ellipse cut from the cylinder $x^2+y^2=4$ by the tilted plane $z=1+2x-y$, traced clockwise viewed from $+z$.

Approach. Apply Stokes' theorem over the flat elliptical patch of the plane bounded by $C$, projected onto the disk $x^2+y^2\leq4$. Compute the curl once, then account for orientation: the standard "upward normal $\leftrightarrow$ counterclockwise-from-above" pairing gives the CCW integral, and the requested CW integral is its negative.

Curl of $\mathbf v$. With $\mathbf v=(z+x,\,-2y,\,y^2)$, $$\nabla\times\mathbf v=\left(\frac{\partial(y^2)}{\partial y}-\frac{\partial(-2y)}{\partial z},\ \frac{\partial(z+x)}{\partial z}-\frac{\partial(y^2)}{\partial x},\ \frac{\partial(-2y)}{\partial x}-\frac{\partial(z+x)}{\partial y}\right)=(2y,\ 1,\ 0).$$
Upward-oriented surface element. For $z=f(x,y)=1+2x-y$, 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 and integrate over the disk (CCW case first). $$(\nabla\times\mathbf v)\cdot(-2,1,1)=2y(-2)+1(1)+0=1-4y.$$ Over the disk $x^2+y^2\leq4$ (polar, $r\in[0,2]$), $\iint(1-4y)\,dA$: the $-4y$ term integrates to zero by symmetry, leaving just the area, $\pi(2)^2=4\pi$: $$\oint_{C,\,\text{CCW from }+z}\mathbf v\cdot d\mathbf r=4\pi.$$
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: $$\oint_{C}\mathbf v\cdot d\mathbf r=-4\pi.$$

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

Quantity	Result
$\nabla\times\mathbf v$	$(2y,1,0)$
CCW-from-$+z$ integral	$4\pi$
Requested (CW) integral	$-4\pi$

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: curling the right-hand fingers along the traversal direction, the thumb gives the paired normal.

Why this works. Choosing the flat planar patch (rather than some other capping surface) makes the surface integral a simple double integral over the projected disk. Because the problem specifies clockwise-from-$+z$ — the reverse of the "upward normal" pairing used in the standard formula $d\mathbf S=(-f_x,-f_y,1)dA$ — the computed value must be negated; skipping this step is the single most common error in this class of problem.

Common pitfall. Forgetting to negate for the clockwise orientation, or negating when the curve was actually specified counterclockwise. Also, sign errors in the two curl components that involve $\partial/\partial z$ of the plane's $z=f(x,y)$ substitution.

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

Question 6 of 8: Line Integral via Stokes' Theorem

Question 6: Line Integral via Stokes' Theorem (20 marks)