Been asked to use Stokes' theorem to solve the integral:
$\int _C x dx + (x - 2yz)dy + (x^2 + z)dz $ where C is the intersection between $x^2+y^2+z^2=1$ and $x^2+y^2=x$ and the half space $z>0$.
Just really not sure how to tackle this or how to solve it.
On
Warning: This solution uses Stoke's theorem in language of differential forms like $$\int\limits_{\partial A} \omega = \int\limits_A {d\omega }$$
$\partial A = C$ is the bounding curve of an surface-area say $A$ given by:
$$\begin{array}{l} {x^2} + {y^2} + {z^2} = 1\\ {x^2} + {y^2} = x\\ z > 0 \end{array}$$
Also given a 1-Form $$\omega = xdx + (x - 2yz)dy + ({x^2} + z)dz$$ on the boundary.
For exterior derivative we get
$$d\omega = dx \wedge dy + 2(xdx + ydy) \wedge dz$$
Because $$ {x^2} + {y^2} + {z^2} = 1 $$ observe that:
$$xdx + ydy + zdz = 0$$
and
$$\begin{array}{l} d\omega = dx \wedge dy + 2(xdx + ydy) \wedge dz\\ = dx \wedge dy + 2(xdx + ydy + zdz - zdz) \wedge dz\\ = dx \wedge dy \end{array}$$
Apply now RHS of Stoke's theorem. To calculate RHS, we choose polar-coordinates as follows:
$$\left. \begin{gathered} x = r\cos (\varphi ) \hfill \\ y = r\sin (\varphi ) \hfill \\ \end{gathered} \right\}dx \wedge dy = \left| {\begin{array}{*{20}{c}} {\cos (\varphi )}&{\sin (\varphi )} \\ { - r\sin (\varphi )}&{r\cos (\varphi )} \end{array}} \right|dr \wedge d\varphi = rdr \wedge d\varphi$$
and we are done very quickly:
$$\int\limits_A {dx \wedge dy} = \int\limits_B {rdr \wedge d\varphi } = \int\limits_0^{2\pi } {\int\limits_0^{\frac{1}{2}} {rdrd\varphi } } = 2\pi \cdot \frac{1}{8} = \frac{\pi }{4}$$
Why would anyone compute this integral ($=:J$) using Stokes' theorem? The curve in question goes around the cylinder $\bigl(x-{1\over2}\bigr)^2+y^2={1\over4}$ and stays all the time on the upper half of $x^2+y^2+z^2=1$. It follows that a parametric representation of $C$ (we are free to choose its sense of direction) is given by $$t\mapsto\left\{\eqalign{x(t)&={1\over2}+{1\over2}\cos t \cr y(t)&={1\over2}\sin t\cr z(t)&=\sqrt{1-x^2(t)-y^2(t)}=\sqrt{1-x(t)}=\sin{t\over2}\cr}\right.\qquad(0\leq t\leq2\pi)\ .$$ As $\int_Cx\>dx=\int_Cz\>dz=0$ it remains to compute $$J=\int_C(x-2yz)\>dy+\int_C x^2\>dz=\int_0^{2\pi}\bigl((x(t)-2y(t)z(t))y'(t)+x^2(t)z'(t)\bigr)\>dt\ .$$ Mathematica obtained $J={\pi\over4}$.