Verify Stokes' theorem for $F=2yzi-(x+3y-2)j+(x^2+z)k$ where $S\subset \mathbb{R}^3$ be the surface of intersection of the cylinders $x^2 + z^2 = a^2$ and $x^2 + y^2 =a^2$ in the first octant.
I have evaluated the line integral $\int_C F.dr=-\frac{a^2}{12}(3\pi+8a)$ where C is the boundary of the surface S with four segments as shown in the fig with arrows.
I want to find the surface integral $\int\int_S curl\,F\cdot n\, dS$
$curl F=-2(x-y)j-(1-2z)k$, $n=\frac{yj+zk}{a}$, $dS=\frac{a}{z}dxdy$,
I am unable to solve $\int\int_S curl\,F\cdot n\, dS$. Please help me to solve it.
Best Answer
Please note that at the intersection of both surfaces, $x^2 + y^2 = x^2 + z^2 \implies y = z$
Surface $x^2 + y^2 = a^2$ should be parametrized as,
$ \phi(z, t) = (a \cos t, a \sin t, z)$ with normal vector $ \vec n_1 = (a \cos t, a \sin t, 0)$
There is also a small mistake in your curl. It should be, $\nabla \times \vec F = \langle 0, 2 (y - x), - (1 + 2z) \rangle$
The surface integral is,
$ \displaystyle \int_0^{\pi/2}\int_0^{a \sin t} 2a^2 \sin t ~(\sin t - \cos t) ~ dz ~ dt $
Similarly surface $x^2 + z^2 = a^2$ can be parametrized as $(a \sin t, y, a \cos t)$ with normal vector $ \vec n_2 = (a \sin t, 0, a \cos t)$
The surface integral is,
$ \displaystyle \int_0^{\pi/2}\int_0^{a \cos t} - a \cos t ~(1 + 2 a \cos t) ~ dy ~ dt $
Adding both, we do get $ ~~\displaystyle -\frac{a^2}{12}(3\pi+8a)$.