Line Integral via Stokes' Theorem (Cylinder ∩ Plane): 04-BS-1, December 2018, Q8

Question 8: Line Integral via Stokes' Theorem (Cylinder ∩ 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. $\mathbf v=(4z,\,-2x,\,2x)$; $C$ = the ellipse cut from the cylinder $x^2+y^2=1$ by the plane $z=1+y$, traversed clockwise viewed from $+z$.

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

Approach. Apply Stokes' theorem over the flat elliptical patch $S$ (the plane $z=1+y$ restricted to $x^2+y^2\le1$) bounded by $C$. Compute the constant curl, dot it with the constant surface-element vector from the natural parametrization, then correct for orientation.

The ellipse $C$ (blue) cut from the cylinder $x^2+y^2=1$ (dashed base circle) by the tilted plane $z=1+y$.

Curl of $\mathbf v$. $$\operatorname{curl}\mathbf v=\left(\frac{\partial(2x)}{\partial y}-\frac{\partial(-2x)}{\partial z},\ \frac{\partial(4z)}{\partial z}-\frac{\partial(2x)}{\partial x},\ \frac{\partial(-2x)}{\partial x}-\frac{\partial(4z)}{\partial y}\right)=(0,\,2,\,-2).$$
Surface-element vector. Parametrize the plane patch as $\mathbf r(x,y)=(x,y,1+y)$, so $\mathbf r_x=(1,0,0)$, $\mathbf r_y=(0,1,1)$, and $$\mathbf r_x\times\mathbf r_y=(0,-1,1)$$ which is the upward (CCW-from-$+z$) normal by the standard right-hand-rule convention for this parametrization.
Dot and integrate (CCW value). $\operatorname{curl}\mathbf v\cdot(\mathbf r_x\times\mathbf r_y)=(0)(0)+(2)(-1)+(-2)(1)=-4$, a constant, so $$\oint_{C,\,\text{CCW}}\mathbf v\cdot d\mathbf r=\iint_{x^2+y^2\le1}(-4)\,dA=-4\cdot\pi(1)^2=-4\pi.$$
Flip for the requested (clockwise) orientation. The question specifies clockwise viewed from above, i.e. the reverse of the CCW value just found: $$\oint_{C,\,\text{CW}}\mathbf v\cdot d\mathbf r=-(-4\pi)=4\pi.$$

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

Quantity	Result
$\operatorname{curl}\mathbf v$	$(0,2,-2)$
$\mathbf r_x\times\mathbf r_y$	$(0,-1,1)$
CCW-from-above value	$-4\pi$
CW-from-above value (requested)	$4\pi$

Concept & Theory

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

Key relations. $\displaystyle\oint_C\mathbf v\cdot d\mathbf r=\iint_S(\nabla\times\mathbf v)\cdot d\mathbf S$, where $S$ is any surface bounded by $C$ and the orientation of $\partial S=C$ follows the right-hand rule from $S$'s chosen normal.

Why this works. The flat elliptical patch cut from the plane by the cylinder is the simplest surface bounded by $C$; because $\operatorname{curl}\mathbf v$ and the surface-element vector are both constant over this patch, the surface integral collapses to (constant) $\times$ (area) with no need to integrate a varying integrand.

Common pitfall. Using the CCW-from-above value directly without checking the problem's stated orientation — here the requested clockwise traversal is the opposite of the natural $\mathbf r_x\times\mathbf r_y$ normal's boundary orientation, so the sign must be flipped; always re-derive this per problem rather than assuming a fixed sign convention.

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

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

Question 8: Line Integral via Stokes' Theorem (Cylinder ∩ Plane) (20 marks)