# Line integral of the vector field $(x-y,y-z,x+z)$ along a path on the unit sphere from $(1,0,0)$ to $(0,1,0)$ to $(0,0,1)$ back to $(1,0,0)$

I'm asked to calculate the line integral of the vector field $$K=(x-y,y-z,x+z)$$ along a path on the unit sphere centered at $$(0,0,0)$$. The path first goes from $$(1,0,0)$$ to $$(0,1,0)$$ (section (1)), then from $$(0,1,0)$$ to $$(0,0,1)$$ (section(2)), the from $$(0,0,1)$$ back to $$(1,0,0)$$ (section(3)).

I have no solution to this old exam question, so I would like that you check my calculations. Thank you.

So I split the integral into 3 parts.

(1) : Here, $$z=0$$. We can parametrize the path with $$(\cos(\theta), \sin(\theta), 0)$$ where $$\theta \in [0, \frac{\pi}{2}]$$.
The derivative is $$(-\sin(\theta),cos(\theta),0)$$
(2) : Here, $$x=0$$. We can parametrize the path with $$(0, \cos(\theta), \sin(\theta))$$ where $$\theta \in [0, \frac{\pi}{2}]$$.
The derivative is $$(0,-\sin(\theta),cos(\theta))$$
(3): Here, $$y=0$$. We can parametrize the path with $$(\sin(\theta),0,\cos(\theta))$$ where $$\theta \in [0, \frac{\pi}{2}]$$.
The derivative is $$(\cos(\theta),0,-\sin(\theta))$$

Each time,we have the integral from $$0$$ to $$\frac{\pi}{2}$$ with respect to $$\theta$$

(1) : $$\int_{0}^{\frac{\pi}{2}}(\cos(\theta)-\sin(\theta),\sin(\theta)-0,\cos(\theta)+0)\cdotp (-\sin(\theta),cos(\theta),0) d\theta$$
(2) : $$\int_{0}^{\frac{\pi}{2}}(0- \cos(\theta),\cos(\theta)-\sin(\theta),0+\sin(\theta))\cdotp (0,-\sin(\theta),cos(\theta)) d\theta$$
(3) : $$\int_{0}^{\frac{\pi}{2}}(\sin(\theta)-0,0-\cos(\theta),\sin(\theta)+\cos(\theta))\cdotp (\cos(\theta),0,-\sin(\theta)) d\theta$$

So we get : $$\int_{0}^{\frac{\pi}{2}} -\sin(\theta)\cos(\theta) + \sin^2(\theta)+\cos(\theta)\sin(\theta) d\theta$$$$+ \int_{0}^{\frac{\pi}{2}} -\sin(\theta)\cos(\theta) + \sin^2(\theta)+\cos(\theta)\sin(\theta) d\theta$$ $$+ \int_{0}^{\frac{\pi}{2}} \sin(\theta)\cos(\theta) - \sin^2(\theta)-\cos(\theta)\sin(\theta) d\theta$$ $$= \int_{0}^{\frac{\pi}{2}}\sin^2(\theta) d\theta=\frac{\pi}{4}$$

So the result seems to make sense to me. I mean it's not an ugly result but still, I have no idea if it's correct.

Also, we could of course use Stoke's theorem in this question, finding $$rot(K)$$ and the cross product of the two partial derivatives of some parametrization, and then integrating over the entire region, but do you agree that this is much more complicated than if we simply evaluate the three different paths ?

If you want to check it vs Stokes

$$\nabla \times F = (1,-1,1)$$

$$n = (x,y,z)$$

Want to do this in spherical?

$$\int_0^{\frac {\pi}{2}}\int_0^{\frac {\pi}{2}} (\cos\theta\sin\phi - \sin\theta\sin\phi + \cos\phi)(\sin \phi)\ d\phi\ d\theta$$

$$\frac {\pi}{4}$$

• Just a question. Where did get. $\cos(\theta) \sin(\phi))$ and the next one ? If you take the dot product of n with the rotation, dont you simply get $x-y+z$ ? Jan 8, 2019 at 21:27
• Ah okay, nevermind. It’s spherical coordinates, not polar coordinates. Sorry haha Jan 8, 2019 at 21:30