Evaluate the line integral $\int_\gamma \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy + \ln{(z^4+z^2+1)}dz$ using Stokes' Theorem

Question

I found the following problem in a textbook (translated):

Evaluate $$\int_\gamma \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy + \ln{(z^4+z^2+1)}dz$$ where $\;\gamma\;$ is given by the intersection between the cone $x^2+y^2=z^2$, and the plane $x+y+10z=10$. The orientation is chosen to be so that the curve goes one revolution around the $z$-axis clockwise (seen from above)

I am thinking to use stokes theorem. I evaluate the curl to be $(0,0,0)$ and the line integral becomses $0$. I know this is incorrect, what am I missing? I also see that the vector field is undefined along the $z$-axis.

Answer 1

As the vector field is not defined at any point on z-axis, it will be easiest to choose surface of the cone between $z = 1 - \frac{1}{10} (x + y)$ and say $z = 2$ [any value of $z (\ne 0)$ which intersects the cone and is clear from the other plane $x+y+10z = 10$ within the bounds of the cone]. Now our surface has two boundaries.

So by Stokes theorem we have,

$\displaystyle \iint_S (\nabla \times \vec F) \cdot \hat {n} \ dS = 0 = \int_{C2} \vec F \ dr - \int_{C1} \vec F \ dr$

$\displaystyle - \int_{C1} \vec F \ dr = - \int_{C2} \vec F \ dr$

where $C1$ is the curve given by intersection of $x+y+10z = 10$ and $z^2 = x^2+y^2$. $C2$ is intersection of plane $z = 2, z^2 = x^2+y^2$.

We parametrize curve $C2$ as $ \ r(t) = (2 \cos t, 2 \sin t, 2)$

$r'(t) = (-2\sin t, 2 \cos t, 0)$

$\vec F (r(t))= (- \frac{2 \sin t}{4}, \frac{2 \cos t}{4}, \ln (21))$

$\displaystyle - \int_{C2} \vec F \ dr = - \int_0^{2\pi} \vec F(r(t)) \cdot r'(t) \ dt = - 2\pi$

Answer 2

As @RobertTheTutor notes, you need a different strategy. In fact you thought of one, but let me show how it works. Give $\gamma$ the parameterization$$x=r\cos\varphi,\,y=r\sin\varphi,\,\pm r=z=1-r(\cos\varphi+\sin\varphi)/10$$so $r=\frac{\pm10}{10\pm(\cos\varphi+\sin\varphi)}$. In fact the $\pm$ must be $+$ since $r\ge0$, so$$x=\frac{10\cos\varphi}{10+\cos\varphi+\sin\varphi},\,y=\frac{10\sin\varphi}{10+\cos\varphi+\sin\varphi},\,z=\frac{10}{10+\cos\varphi+\sin\varphi}.$$Hence $\frac{xdy-ydx}{x^2+y^2}+\ln(z^4+z^2+1)dz$ is $d\varphi$ times a function of $\varphi$ you can integrate from $2\pi$ to $0$ to get the desired orientation (if I've read the instructions correctly). The $z$ part will of course just give us $0$ by periodicity, but the first term is just $d\varphi$, making the answer $-2\pi$. That the result is a multiple of $2\pi$ won't surprise anyone familiar with the effect poles have on contour integrals in complex analysis.

Answer 3

Since the vector field is not defined on the entire surface, we cannot use Stokes' Theorem, so we must perform the line integral directly.

We parametrize the curve: $0 \leq t \leq 2\pi$, with $x = \cos t, y = \sin t, z = 1-\frac{1}{10}(\cos t + \sin t)$

$$\int \overrightarrow F \cdot d\overrightarrow r = \int_0^{2\pi} \frac{-\sin t}{\sin^2 t + \cos^2 t}\cdot(-\sin t) +\frac{\cos t}{\sin^2 t + \cos^2 t}\cdot(\cos t) dt + \int\ln (z^4 + z^2 + 1)dz $$ which simplifies to $$2\pi + \int\ln (z^4 + z^2 + 1)dz $$ where $z = 1-\frac{1}{10}(\cos t + \sin t)$.

Integrating by parts, I find $$\int \ln (z^4 + z^2 + 1) dz = z\ln(z^4 + z^2 + 1)|_{t=0}^{t=2\pi} - \int \frac{z(4z^3+2z)}{z^4+z^2 + 1}dz = - \int \frac{z(4z^3+2z)}{z^4+z^2 + 1}dz $$ from here it looks like a straightforward slog through long division and partial fractions, since $z^4 + z^2 + 1 = z^4 + 2z^2 + 1 - z^2 = (z^2 + 1)^2 - z^2 = (z^2 + 1 + z)(z^2 + 1 -z)$.

EDIT: Actually, since $z$ is a periodic function of $t$ with period $2\pi$, integrating from $0$ to $2\pi$ $dt$ should give $0$ for the entire $z$ integral, leaving the answer $2\pi$. EDIT2: And don't forget a minus sign because it is clockwise, not counterclockwise, so the answer is $-2\pi$.