It is given vector field $v(x,y,z)=\begin{pmatrix} -x^3-2y \\ 3y^5z^6 \\ 3y^6z^5-z^4 \end{pmatrix} $
Evaluate the surface integral over the hemisphere F, which is defined by $z≥0$ and $x^2 + y^2 + z^2 = 4$ Integral is defined as:
$\int_{\partial F}^{} \! v \cdot \, dx $
What I did:
I used spherical coordinates and I calculated $\nabla \times v=(0,0,2)$
Now I wanted to calculate the following integral,
$\int_{0}^{2} \! \int_{0}^{\frac{\pi }{2} } \! \int_{0}^{2\pi } \! r^2sin\theta (0,0,2) \, d\theta \, d\varphi \, dr$
But I dont get the right solution. Can someone help me with this? P.S. I am also confused why is in the integral, they gave, $dx$?
Let me clarify the notations for you: $\int_{\partial F} v\cdot dx$ should be $\int_{\partial F} \mathbf{v}\cdot d\mathbf{x}$, where $\mathbf{v}=(-x^3-2y,3y^5z^6,3y^6z^5-z^4)$ and $d\mathbf{x}=(dx,dy,dz)$. The integral, in its clearest form, is: $$\int_{\partial F} (-x^3-2y)dx+(3y^5z^6)dy+(3y^6z^5-z^4)dz.$$
Solution I (Stokes' theorem): $$\int_{\partial F} \mathbf{v}\cdot d\mathbf{x}=\int_F \nabla\times \mathbf{v}\cdot n\,dS,$$ where $\nabla \times \mathbf{v}=(0,0,2)$ and $n$ is the unit outward-pointing normal to the surface $n=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}=\frac{1}{2}(x,y,z)$.
Use the parametrisation $\varphi:U=\{(x,y)\mid x^2+y^2\leq 4\}\to F$, given by $\varphi(x,y)=(x,y,\sqrt{4-x^2-y^2})$. Then $$D\varphi(x,y)=\begin{pmatrix}1 & 0 \\ 0 & 1\\ \frac{-x}{\sqrt{4-x^2-y^2}} & \frac{-y}{\sqrt{4-x^2-y^2}}\end{pmatrix}=:(v_1\,\, v_2)$$ We then find $$dS(v_1,v_2)=\det(n,v_1,v_2)=\frac{1}{2}\det\begin{pmatrix}x & 1 & 0 \\ y & 0 & 1\\ \sqrt{4-x^2-y^2} & \frac{-x}{\sqrt{4-x^2-y^2}} & \frac{-y}{\sqrt{4-x^2-y^2}}\end{pmatrix}=\frac{2}{\sqrt{4-x^2-y^2}}.$$ Finally, $$\begin{align}\int_{\partial F} \mathbf{v}\cdot d\mathbf{x} &=\int_F \nabla\times \mathbf{v}\cdot n\,dS\\ &=\int_U [(0,0,2)\cdot \frac{1}{2}(x,y,\sqrt{4-x^2-y^2})] \frac{2}{\sqrt{4-x^2-y^2}}\\ &=2\int_U 1\\ &=2\,\text{area}(U)\\ &=2\pi (2^2)\\ &=8\pi.\end{align}$$
Solution II: Note that on the boundary $\partial F=\{(x,y)\mid x^2+y^2=4\,\text{and}\, z=0\}$, the integral reduces to $$\int_{\partial F} (-x^3-2y)dx.$$ Use the parametrisation $\phi(t):[0,2\pi]\to \partial F$ given by $\phi(t)=(2\cos t,2\sin t)$. Then $D\phi(t)=(-2\sin t,2\cos t)$ and $dx=-2\sin t\, dt$. Hence the integral is $$\int_0^{2\pi} (-(2\cos t)^3-2(2\sin t))(-2\sin t)\,dt=8\pi.$$
Solution III (Stokes' theorem with differential forms): Use the same parametrisation $\varphi$ as in solution I. Given the vector field $\mathbf{v}$, there is an associated differential form $$\omega=(-x^3-2y)dx+(3y^5z^6)dy+(3y^6z^5-z^4)dz.$$ $$\begin{align}\int_{\partial F} \mathbf{v}\cdot d\mathbf{x} &=\int_F \nabla\times \mathbf{v}\cdot n\,dS\\ &=\int_F d\omega\\ &=\int_U 2dx\wedge dy\\ &=\int_U 2\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}\\ &=\int_U 2\\ &=2\,\text{area}(U)\\ &=8\pi.\end{align}$$
Solution IV: Let $D=\{(x,y)\mid x^2+y^2\leq 4\}$, $C=\{(x,y)\mid x^2+y^2=4\}$. Then $C=\partial D=\partial F$. $$\int_{C=\partial F=\partial D} \mathbf{v}\cdot d\mathbf{x}=\int_D \nabla\times \mathbf{v}\cdot n\,dS=\int_D (0,0,2)\cdot (0,0,1)\,dS=2\,\text{area}(D)=8\pi$$ since $(0,0,1)$ is a unit normal to $D$.
To sum up, the above solutions follow from the following identities, each of which is a special case of the generalised Stokes' theorem. $$\begin{align}\int_{\partial F} \mathbf{v}\cdot d\mathbf{x}&\stackrel{I,IV}{=}\int_F \nabla\times \mathbf{v}\cdot n\,dS\\ &\stackrel{II}{=}\int_{\partial F} \omega\\ &\stackrel{III}{=}\int_F d\omega \end{align}$$