Question 7 of 8: Line Integral via Stokes' Theorem (Cylinder ∩ Tilted Plane)

Question 7: 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 $x^2+y^2=4$ (radius 2) with the plane $z=3-2x+y$, traversed counterclockwise viewed from $+z$. Vector field $\mathbf v=(x,\ x-y,\ yz)$.

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

Approach. Apply Stokes' theorem over the flat elliptical patch of the plane bounded by $C$, projected onto the disk $x^2+y^2\leq4$. The requested orientation (CCW from $+z$) is exactly the one paired with an upward-pointing normal by the right-hand rule, so no sign flip is needed here.

1. Curl of $\mathbf v$. With $\mathbf v=(x,\,x-y,\,yz)$, $$\nabla\times\mathbf v=\left(\frac{\partial(yz)}{\partial y}-\frac{\partial(x-y)}{\partial z},\ \frac{\partial(x)}{\partial z}-\frac{\partial(yz)}{\partial x},\ \frac{\partial(x-y)}{\partial x}-\frac{\partial(x)}{\partial y}\right)=(z,\ 0,\ 1).$$
2. Upward-oriented surface element. For $z=f(x,y)=3-2x+y$, the upward-normal surface element (paired with CCW-from-above) is $d\mathbf S=(-f_x,-f_y,1)\,dx\,dy=(2,-1,1)\,dx\,dy$.
3. Dot the curl with the surface element. On the surface, $z=3-2x+y$, so $$(\nabla\times\mathbf v)\cdot(2,-1,1)=2z+0(-1)+1=2(3-2x+y)+1=7-4x+2y.$$
4. 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, leaving only the constant term times the area $\pi(2)^2=4\pi$: $$\oint_C\mathbf v\cdot d\mathbf r=\iint_{x^2+y^2\le4}(7-4x+2y)\,dA=7(4\pi)=28\pi.$$

$$\oint_C\mathbf v\cdot d\mathbf r=\boxed{28\pi}\approx87.96$$