# Stokes' theorem - circle cross section in a sphere

A circle C is cut on the surface of the hemisphere $x^2 + y^2 + z^2 = 1,z≥0$ by the cylinder $x^2 + y^2 = y.$Evaluate $\int_{C} -y^2\,dx +y^2\,dy+z^2\,dz$ where the direction round C is such that the point (0,0,1) is directed into the first octant.

Attempt: So completing the square gives a cylinder of centre (0,1/2) and radius 1/2. Using Stoke's thm, I identified the vector field F to be $−y^2 i + y^2j +z^2 k$ and took the curl of it to give $2yk$. I believe everything is right up to here.

I am confused about what the surface is here that is bounded by C. I realise that to compute $d\vec{S} =\frac{r_u \times r_v}{|r_u \times r_v|}dS,$ I have to find a suitable parametrisation of some surface. I found where the cylinder and sphere intersected :$y+z^2 = 1=>z=\sqrt{1−y}$ since $z≥0$.So then my parametrisation would be $r(x,y)=xi+yj+\sqrt{1-y} k$ from which I could then compute two tangent vectors and a normal.

I am not sure if my parametrisation is correct. Can anyone offer any advice? Many thanks

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Be careful here! The intersection of the cylinder and hemispherenis far from a circle! –  Ted Shifrin Jun 5 '13 at 23:47
@TedShifrin good point. That mistake in his first reply slipped past my notice. –  James S. Cook Jun 6 '13 at 13:32

Your parametrization ought to have $x^2+y^2+z^2=1$ yet $x^2+y^2+(\sqrt{1-y})^2=x^2+y^2+1-y$. This expression does reduce to $1$ when we set $x^2+y^2=y$. However, this means the parametrization is only on the sphere at the boundary of the subset of the sphere you are studying. In other words, you have parametrized the cylinder, not the sphere.
The sphere, has $\vec{r}(x,y) = < x,y, \sqrt{1-x^2-y^2}>$ in the region of space where $z>0$. However, I much prefer spherical coordinate-based parametrizations when I can use them, but here the off-center cylinder seems to discourage such a practice. Hope this helps.
Correct, the cylinder's interior as it intersects the $xy$-plane shows you the natural domain. You want to integrate on $x^2+(y-\frac{1}{2})^2 \leq \frac{1}{4}$. Or, change to polars $x = r\cos t, y = \frac{1}{2}+r\sin t$ for $0 \leq r \leq 1/2$ and $0 \leq t \leq 2\pi$. That'll give you a nicer integration. –  James S. Cook Nov 10 '12 at 19:46
Many thanks. I have found a normal and then said $curl\vec{F} \cdot \vec{n} dS = 2y\,dy\,dx.$ I then change this to polar coordinates to give $2(\frac{1}{2} + r\sin t)(-r\sin t)(r\cos t) dt$ integrated over $r \in\,[0,1/2], t \in\,[0,2\pi]$? –  CAF Nov 10 '12 at 20:47
@JamesS.Cook: You wrote that "in other words, you have parametrized the cylinder, not the sphere." Since $$y = z^2 + 1 \Rightarrow z = \sqrt{1 - y} = \text{the curve of intersection of the cylinder and sphere},$$ shouldn't $$(x, y,\sqrt{1 - y}) = \text{parameterisation of the intersection of the cylinder and sphere}?$$ How is it the paramterisation of the cylinder? –  LePressentiment May 12 '13 at 13:47