a) Let $S$ the surface $4x^2+9y^2+36z^2=36$, $z \ge 0$. Let $\vec{F}=y\vec{i}+x^2\vec{j}+(x^2+y^4)^{3/2}\sin(e^{xyz})\vec{k}.$ Calculate the integral $\iint (\text{curl }\vec{F})\cdot\vec{n}\,dS$.
b) Calculate the same integral when $S$ is the whole surface of the ellipsoid.
My solution: $\vec{F}=(F_1,F_2,F_3),\ F_1(x,y,z)=y,\ F_2(x,y,z)=x^2$ $\vec{σ}:[0,2\pi]\rightarrow \mathbb{R^2},\ \vec{σ}=(x(t),y(t)),\ x(t)=3cost,\ y(t)=2sint$ $\iint (\text{curl }\vec{F})\cdot\vec{n}\,dS=\int_{θS}\vec{F}\cdot d\vec{s}=\int_{C^+}F_1dx+F_2dy=\int_0^{2\pi}(F_1\frac{dx}{dt}+F_2\frac{dy}{dt})dt=\int_0^{2\pi} y\frac{dx}{dt}+x^2\frac{dy}{dt}=\int_0^{2\pi}[2sint(-3sint)+(3cost)^22cost]dt=-6\pi$ Is there any other solution?
Since $\vec\nabla\cdot\left(\vec\nabla\times\vec F\right)=0$, the integral over any closed surface $$\oint\left(\vec\nabla\times\vec F\right)\cdot d^2\vec A=0$$ By the divergence theorem. That resolves part b already. For part a, we have $$\int\int_{S_1}\left(\vec\nabla\times\vec F\right)\cdot d^2\vec A+\int\int_{S_2}\left(\vec\nabla\times\vec F\right)\cdot d^2\vec A=0$$ Where $S_1$ is the given surface and $S_2$ is the part of the $xy$-plane with $$\frac{x^2}{3^2}+\frac{y^2}{2^2}\le1$$ With the positive sense of the surface being directed downwards, $$d^2\vec A=-\hat k\,dx\,dy$$ So we only need the $z$-component to do the integral, and $$\int\int_{S_2}\left(\vec\nabla\times\vec F\right)\cdot d^2\vec A=\int\int_{S_2}(2x-1)(-d^2A)=\pi\cdot3\cdot2$$ Where we have used the mean value of $x$ of $0$ over the ellipse and the mensuration formula for the area of an ellipse. Thus the answer in part a is $$\int\int_{S_1}\left(\vec\nabla\times\vec F\right)\cdot d^2\vec A=-6\pi$$