Line Integral via Stokes' Theorem: 04-BS-1, May 2013, Q7

Question 7: 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 cylinder $x^2+y^2=4$ (radius 2) and plane $z=3-2x+y$, traversed CCW viewed from $+z$; vector field $\mathbf v=(x,\ x-y,\ yz)$.

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

Figure — closed curve C = intersection of the cylinder x²+y²=4 with the tilted plane z+2x−y=3, traversed CCW as viewed from +z; Stokes converts ∮_C v·dr to a flat disk integral.

Approach. $C$ is a closed curve, so apply Stokes' theorem: $\oint_C\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS$ over the flat elliptical cap $S$ (the plane patch inside the cylinder), with $\mathbf n$ chosen upward to match the CCW-from-above orientation.

Compute the curl. $$\nabla\times\mathbf v=\left(\frac{\partial(yz)}{\partial y}-\frac{\partial(x-y)}{\partial z},\ \frac{\partial x}{\partial z}-\frac{\partial(yz)}{\partial x},\ \frac{\partial(x-y)}{\partial x}-\frac{\partial x}{\partial y}\right)=(z,\ 0,\ 1).$$
Set up the flat-cap integral. The plane $2x-y+z=3$ has normal $\mathbf n\propto(2,-1,1)$, whose $z$-component is positive — matching the CCW-from-$+z$ orientation. Using $\mathbf n\,dS=(2,-1,1)\,dx\,dy$ (projected onto the $xy$-disk): $$(\nabla\times\mathbf v)\cdot\mathbf n = (z,0,1)\cdot(2,-1,1)=2z+1.$$ On the plane, $z=3-2x+y$, so $(\nabla\times\mathbf v)\cdot\mathbf n=2(3-2x+y)+1=7-4x+2y$.
Integrate over the disk $x^2+y^2\le4$. By symmetry, $\iint x\,dA=\iint y\,dA=0$ over a disk centred at the origin, so only the constant term survives: $$\oint_C\mathbf v\cdot d\mathbf r=\iint_{x^2+y^2\le4}(7-4x+2y)\,dA = 7\cdot(\pi\cdot2^2) = 28\pi.$$

$$\oint_C \mathbf v\cdot d\mathbf r=\boxed{28\pi\approx 87.96}$$

Quantity	Result
$\oint_C \mathbf v\cdot d\mathbf r$	$28\pi\approx 87.96$

Concept & Theory

Topic. Stokes' theorem — converting a 3-D closed-curve line integral into a surface integral.

Key relations. $\oint_C\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot\mathbf n\,dS$; for $\mathbf v=(P,Q,R)$, $\nabla\times\mathbf v=(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y)$; for a plane $ax+by+cz=d$, $\mathbf n\,dS=(a,b,c)\,dx\,dy$ (up to orientation and normalization) when projecting onto the $xy$-plane.

Why this works. $C$ bounds infinitely many surfaces, but Stokes' theorem lets us pick the most convenient one — here the flat elliptical cap cut from the tilted plane by the cylinder — because the theorem's result is independent of which capping surface is chosen (as long as orientation is consistent). Using the flat cap converts the awkward tilted-curve line integral into a routine double integral over a disk, where symmetry kills the linear terms immediately.

Common pitfall. Choosing the wrong sign of $\mathbf n$ (must have positive $z$-component to match "CCW viewed from $+z$"); forgetting that $dS$ over a tilted plane projected to the $xy$-disk absorbs the plane's normal vector directly (no extra $\sec\gamma$ factor needed when using $\mathbf n\,dS=(a,b,c)\,dA$ directly, since the $(a,b,c)$ vector already has the right magnitude relationship to the projection).

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 9 (Vector Differential Calculus, §9.9 Stokes' Theorem).

Question 7 of 8: Line Integral via Stokes' Theorem

Question 7: Line Integral via Stokes' Theorem (20 marks)