Let $C$ be the intersection curve between the plane $z = 10 – x – y$ and the cylinder $x^2+y^2 = 1$, oriented such that the projection of the curve onto the xy-plane is positively oriented. Determine the work done by the force field $\mathbf{F} = (x, x^3, z^3)$ in circulating around the curve $C$.
Long story short I parameterized the curve using polar coordinate:
$$r(t,r) = (r\cos(t), r\sin(t), 10 – r\cos(t) – r\sin(t))$$
Then I partial differentiated with respect to t and r and then took their cross product to get the normal vector:
$$r_{t} \times r_{r} = (-r, -r, -r)$$
The curl of $F$ became $F = (0,0, 3x^2) = (0,0, 3r^2\cos^2(t))$ and the dot product between the normal and $F$ is then $-3r^3\cos^2(t)$.
$$\iint_{D} -3r^3\cos^2(t) = -\frac{3\pi}4$$
The correct answer is positive, and it got me thinking is because it says in the question that curve projection on the xy-plane has a positive orientation and using the right hand rule, the normal points upwards. This means that I've to change the direction of my normal vector right? Is this the only reason?
While we are on the subject, I don't need to use Jacobian determinant right? As in multiply by "r" when I parametrize it. Is it because I am already calculating the change in the area caused by the variable change when I take the cross product?
Best Answer
Your parametrization of the given curve $C$ is wrong. In the first place because it has two parameters $r$ and $t$ for a $1$-dimensional object (a curve).
Anyway, a positively oriented (counterclockwise) parametrization of the curve $C$ is $$\mathbf{r}(t) = \left(\cos(t),\sin(t), 10-\cos(t)-\sin(t)\right)\quad \text{for $t\in[0,2\pi]$.}$$
A parametrization of the plane surface $S$ such that $\partial S=C$ is $$\mathbf{r}(x,y)=(x,y,10-x-y) \quad\text{with $(x,y)$ such that $x^2+y^2\leq 1$.}$$ By the right hand rule, the induced orientation on $S$ is upwards and $d\mathbf{S} =(1,1,1)$ which is opposite to yours. Hence, applying Stokes' theorem, we find $$\begin{align} \int_C \mathbf{F}\cdot d\mathbf{s}&=\iint_S \text{curl} \mathbf{F}\cdot d\mathbf{S}\\ &=\iint_{x^2+y^2\leq 1}(0,0,3x^2)\cdot (1,1,1) dxdy\\ &=\int_{0}^1\int_0^{2\pi}3r^2\cos^2(t)\, rdrdt=\frac{3\pi}{4}.\end{align}$$