Line Integral via Stokes' Theorem: 04-BS-1, May 2016, Q8

Question 8: 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. A closed curve $C$ (an ellipse) cut from the plane $z=1+y-3x$ by the cylinder $x^2+y^2=9$, and the vector field $\mathbf v=(2z^2,-2y,2y)$.

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

$C$ bounds the flat elliptical disk $S$ cut from the plane by the cylinder; Stokes' theorem converts the line integral to a flux integral of $\operatorname{curl}\mathbf v$ over $S$.

Approach. Apply Stokes' theorem over the planar disk $S:\ z=1+y-3x,\ x^2+y^2\le9$, being careful to orient the surface normal consistently with the stated clockwise-from-$+z$ traversal of $C$.

Compute the curl. For $\mathbf v=(2z^2,-2y,2y)$, $$\operatorname{curl}\mathbf v=\left(\frac{\partial(2y)}{\partial y}-\frac{\partial(-2y)}{\partial z},\ \frac{\partial(2z^2)}{\partial z}-\frac{\partial(2y)}{\partial x},\ \frac{\partial(-2y)}{\partial x}-\frac{\partial(2z^2)}{\partial y}\right)=(2,\,4z,\,0).$$
Parametrize $S$ and find its (upward-normal) orientation. With $\mathbf r(x,y)=(x,y,1+y-3x)$ over the disk $x^2+y^2\le9$, $\mathbf r_x=(1,0,-3)$, $\mathbf r_y=(0,1,1)$, and $$\mathbf r_x\times\mathbf r_y=(0\cdot1-(-3)\cdot1,\ -3\cdot0-1\cdot1,\ 1\cdot1-0\cdot0)=(3,-1,1).$$ This normal has positive $z$-component, so it corresponds (by the right-hand rule) to $C$ traversed counterclockwise viewed from $+z$ — the opposite of the stated clockwise orientation. So the flux integral with this normal must be negated at the end.
Form the flux integrand. Substituting $z=1+y-3x$ into $\operatorname{curl}\mathbf v=(2,4z,0)$ and dotting with $(3,-1,1)$: $$\operatorname{curl}\mathbf v\cdot(\mathbf r_x\times\mathbf r_y)=2(3)+4(1+y-3x)(-1)+0=6-4-4y+12x=2+12x-4y.$$
Integrate over the disk (CCW orientation first). By symmetry of the disk $x^2+y^2\le9$ about the origin, $\iint x\,dA=\iint y\,dA=0$, so only the constant term survives: $$\iint_S\operatorname{curl}\mathbf v\cdot d\mathbf S\Big|_{\text{CCW}}=\iint_{x^2+y^2\le9}2\,dA=2\cdot\pi(3)^2=18\pi.$$
Negate for the stated (clockwise) orientation. $$\oint_C\mathbf v\cdot d\mathbf r=\boxed{-18\pi}\approx-56.55$$ (Verified directly: parametrizing $C$ clockwise as $x=3\cos\theta,\,y=-3\sin\theta$ and numerically integrating $\mathbf v\cdot d\mathbf r$ around the loop reproduces $-18\pi$.)

Quantity	Result
$\operatorname{curl}\mathbf v$	$(2,\,4z,\,0)$
Surface normal $\mathbf r_x\times\mathbf r_y$	$(3,-1,1)$ (upward $\Rightarrow$ CCW from $+z$)
Flux, CCW orientation	$18\pi$
$\oint_C\mathbf v\cdot d\mathbf r$ (clockwise, as given)	$-18\pi\approx-56.55$

Concept & Theory

Topic. Stokes' theorem: converting a closed line integral to a surface flux integral of the curl.

Key relations. $\displaystyle\oint_C\mathbf v\cdot d\mathbf r=\iint_S\operatorname{curl}\mathbf v\cdot\mathbf n\,dS$, where $S$ is any surface bounded by $C$ and $\mathbf n$ is oriented by the right-hand rule relative to the traversal direction of $C$. For a graph surface $z=g(x,y)$, $\mathbf n\,dS=(\mathbf r_x\times\mathbf r_y)\,dA=(-g_x,-g_y,1)\,dA$.

Why this works. The flat elliptical disk cut from the plane is the simplest surface bounded by $C$, avoiding any need to parametrize $C$ itself; because $\operatorname{curl}\mathbf v$ turns out to be affine in $x,y$ after substituting the plane equation, the disk's rotational symmetry kills the linear terms and leaves only the constant term times the disk's area.

Common pitfall. The single most common error here is orientation: computing the flux with the "natural" upward normal from $\mathbf r_x\times\mathbf r_y$ but forgetting that the problem specifies clockwise (viewed from $+z$), which is the reverse of the right-hand-rule default — this flips the sign of the final answer.

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

Question 8 of 8: Line Integral via Stokes' Theorem

Question 8: Line Integral via Stokes' Theorem 20 marks