# Calculating line integral

I'm working on this problem:

Calculate integral $$\int_C\frac{z\arctan(z)}{\sqrt{1+z^2}}\,dz + (y-z^3)\,dx - (2x+z^3)\,dy,$$ where the contour $C$ is defined by equations $$\sqrt{1-x^2-y^2}=z, \quad 4x^2+9y^2 = 1.$$

Seems to me that I know the solution, but I have feeling that I could lost something. Would you help me to clarify this.

First, it is easy to parametrize the contour: $x=\frac{1}{2}\cos\varphi$, $y=\frac{1}{3}\sin\varphi$, $z=\sqrt{1-\frac{\cos^2\varphi}{4}-\frac{\sin^2\varphi}{9}}$ and $\varphi$ goes from $0$ to $2\pi$. So we will have the integral $\int_0^{2\pi}F(\varphi)\,d\varphi$, where the function $F(\varphi)$ is quite complicated.

But I thought about another method. The contour is symmetric and it would provide some simplifications: When variable $z$ goes up and down (on the contour) the value of the function $\frac{z\arctan(z)}{\sqrt{1+z^2}}$ are same in such up-and-down points. So I can write $$\int_C\dots = \int_C + (y-z^3)\,dx - (2x+z^3)\,dy.$$ The same I can conclude for variable $x$ and function $z^3$ and for variable $y$ and function $z^3$. So $$\int_C\dots = \int_C y\,dx - (2x+z^3)\,dy = \int_C y\,dx - 2x\,dy.$$ After that it is much easier to compute the integral using parameterization. $$\int_C y\,dx - 2x\,dy = \int_0^{2\pi}-\frac{1}{6}\sin^2\varphi -2\frac{1}{6}\cos^2\varphi \, d\varphi = -\frac{1}{6}\int_0^{2\pi}1+\cos^2\varphi\,d\varphi = -\frac{\pi}{2}$$

So the answer is $-\frac{\pi}{2}$.

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It's absolutely fine to exploit the symmetries in the given problem. But we need a clear cut argument. Observing that a "variable goes up and down" doesn't suffice.

Relevant are the following symmetries in the parametrization of $C$: $$x(\phi+\pi)=-x(\phi),\quad y(\phi+\pi)=-y(\phi),\quad z(\phi)=z(-\phi)=z(\phi+\pi)\ .$$ This implies $$\dot x(\phi+\pi)=-\dot x(\phi),\quad \dot y(\phi+\pi)=-\dot y(\phi),\quad \dot z(-\phi)=-\dot z(\phi)\ .$$ When computing $W:=\int_C \bigl(P\ dx+Q\ dy+ R\ dz\bigr)$ for $P$, $Q$, $R$ as in your question we therefore immediately get \eqalign{\int_C P\ dx&=\int_0^{2\pi}(y(\phi)-z^3(\phi))\dot x(\phi)\ d\phi=\int_0^{2\pi}y(\phi)\ \dot x(\phi)\ d\phi=-{1\over6}\int_0^{2\pi}\sin^2\phi\ d\phi=-{\pi\over6},\cr \int_C Q\ dy&=\int_0^{2\pi}(-2x(\phi)-z^3(\phi))\dot y(\phi)\ d\phi=-2 \int_0^{2\pi}x(\phi)\ \dot y(\phi)\ d\phi=\ldots=-{\pi\over3},\cr \int_C R\ dz&=\int_{-\pi}^\pi \tilde R(\phi)\dot z(\phi)\ d\phi=0\cr} (where we have used that $\tilde R(-\phi)=\tilde R(\phi)$). It follows that $W=-{\pi\over2}$.

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Thank you for providing clear details! – Nikita Evseev Mar 29 '13 at 5:15

Since I don't want to do these messy calculation by myself, I throw this to SAGE with natural parametrization. Here's the code.

x, y, z, t = var("x, y, z, t")

x = (1/2)*cos(t)
y = (1/3)*sin(t)
z = sqrt(1 - x^2 - y^2)

dx = x.derivative(t)
dy = y.derivative(t)
dz = z.derivative(t)

fx = y - z^3
fy = -(2*x + z^3)
fz = (z*arctan(z))/sqrt(1 + z^2)

df = fx*dx + fy*dy + fz*dz
print N(integral(df(t), (t, 0, 2*pi)))


It returns $-1.5707963267948961 \approx -\pi/2$ so it seems your calculation is right.

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I really liked you approach! Besides it made me to learn what is SAGE. – Nikita Evseev Mar 28 '13 at 12:16