TITLE: Calculate the contour integral for the vector field $z^2 e_1 + x^2 e_2 + y^2 e_3$.
QUESTION [0 upvotes]: The contour that I am given is a circle radius a with centre $(1,1,0)$. So what I thought is to do the parameterisation $\vec x(t)$ = $(1 + a\cos(t), 1 + a\sin(t),  
 0)$ with $t$ going from $0$ to $2\pi$, but this seems to just get me $0$ which I don't think is the right answer.
Any Help?
Edit: so I have that parameterisation. I then did $d\vec x(t)$/dt = $(-asin(t), acos(t), 0)$ 
Now as we have the vector field given above. I times $(1 + acos(t))^2$ with $(acos(t))$ and integrate this between 0 and 2pi which gives me 0.

REPLY [1 votes]: You can solve this exercise by doing the following:
Let's use cylindrical coordinates so that $\gamma (t)=  (1+a\cos(t),1+a\sin(t),0)$, $t\in [0,2\pi)$ is your curve, and $\gamma ' (t) = (-a\sin(t), a\cos(t), 0)$. 
This way you get that $$ \int_{\gamma} \vec{A} \cdot \vec{dl} = \int_0^{2\pi} A(\gamma)\cdot \gamma ' dt =\\ \int_0^{2\pi}\left(0,(1+a\cos(t))^2,(1+a\sin(t))^2\right) \cdot \left((-a\sin(t), a\cos(t), 0)\right) dt=\\ \int_0^{2\pi}(a\cos(t))\cdot(1+a\cos(t))^2  =  2 \pi a^2 $$
(You may find all the steps to solve this last integral on this link)
Note that you can also use Stokes Theorem (a.k.a curl theorem) given that the circle is a closed curve. Therefore, you have:
$$ \int_{\gamma = \delta S} \vec{A} \cdot \vec{dl} = \int_ S (\vec{\nabla} \times\vec{A})\cdot \vec{n} dS$$
Where $S$ is the surface covered by the curve and $\vec{n} = (0,0,1)$ since it's traversed in the positive way - we use the right hand rule -. 
Given that $$\vec{\nabla} \times\vec{A} = (2y,2z,2x) \\(2y,2z,2x)\cdot(0,0,1) = 2x $$
We get $$\int_S 2x dS$$ So, since our curve is $$(x-1)^2 + (y-1)^2 = a^2$$ the integral becomes$$\int_{1-a}^{1+a}2x\left(\int_{-\sqrt{a^2 - (x-1)^2} +1}^{\sqrt{a^2 - (x-1)^2} +1}dy\right)dx = \int_{1-a}^{1+a} 4x\sqrt{a^2 - (-1 + x)^2}dx$$ Solving this integral you should get the same result as before, but as you can see, it is way harder. 
I hope I didn't get any of the calculus wrong.
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