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Evaluate $\displaystyle \int_{C} y^2dx + z^2dy + x^2dz$ where $C$ is the curve of intersection of the sphere $x^2 + y^2 + z^2 = a^{2}$ and the cylinder $x^2 + y^2 = az$ $(a \gt 0, z \ge 0)$ integrated anticlockwise when viewed from origin.

For this line integral I feel that using Stoke's Theorem will be more helpful,but I am not sure how to solve for the normal vector $\hat{n}$. and how to setup the integral

Can anyone tell me how can i setup the integral for Stoke's theorem ?

Thank you.

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    $\begingroup$ $x^2+y^2=az$ is not cylinder. $\endgroup$ – edm Oct 13 at 13:47
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Your second surface is a paraboloid, not a cylinder. Regardless, the trick to Stokes' theorem is that you have the freedom to choose any surface with that curve as a boundary (since there are infinitely many) and you are guaranteed that the integral over all of them will be equal to each other.

In this case, notice that your curve of intersection is contained entirely within the plane $z=\frac{\sqrt{5}-1}{2}a$. So choose to parametrize that plane:

$$\mathbf{r}(x,y) = \left(x,y,\frac{\sqrt{5}-1}{2}a\right), \hspace{20 pt} x^2+y^2\leq \left(\frac{\sqrt{5}-1}{2}\right)^2a^2$$

The normal vector for a plane is easy, it's just $(0,0,1)$ in this case. Which is a bonus because if we're smart about this, we don't have to compute the whole curl, only the $z$ component which is

$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2y$$

And if we're even smarter, we'll recognize that the $z$ component of the curl is an odd function of $y$, and we are integrating over a surface (a disk) with $y$ symmetry, so we can conclude the integral is simply $0$ without parametrizing further.

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