# How Can I use Stoke's Theorem to solve this Line Integral.

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.

• $x^2+y^2=az$ is not cylinder. – edm Oct 13 at 13:47

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.