To evaluate the surface integral $\iint_S(x\,dy\,dz+y\,dx\,dz+z\,dx\,dy)$ How to evaluate the surface integral
$$\iint_S(x\,dy\,dz+y\,dx\,dz+z\,dx\,dy)$$
where $S$ is the outer surface of the ellipsoid
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
that lie above the $xy-$plane.
My thought was to substitute $x=a \sin\theta \cos\psi, y=b \sin\theta \sin\psi, x=c \cos\theta$, where $0\leq \theta\leq \frac{\pi}{2}$ and $0\leq \psi\leq 2\pi$. But in this way i got the answer as $-abc\pi$, which is suppose to be $2abc\pi$. Any help is highly appreciated. Thank you in advance.
My steps
$$\frac{\delta(x,y)}{\delta(\theta,\psi)}=ab\sin\theta \cos\psi$$
$$\frac{\delta(y,z)}{\delta(\theta,\psi)}=bc\sin^2\theta \cos \psi$$
$$\frac{\delta(z,x)}{\delta(\theta,\psi)}=ac\sin^2\theta \sin \psi$$
The given integral
$$=\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}(x\frac{\delta(y,z)}{\delta(\theta,\psi)}+y\frac{\delta(z,x)}{\delta(\theta,\psi)}+z\frac{\delta(x,y)}{\delta(\theta,\psi)})\,d\theta\, d\psi$$
$$=\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}(\sin \theta)\,d\theta\, d\psi$$
$$=-abc\pi$$

By the Gauss-Ostrogradsky divergence theorem the surface integral ∯xdydz+ydxdz+zdxdy is equal to the integral ∫∫∫(div(xi+yj+zk)dxdydz which is equal to ∫∫∫(∂x/∂x+∂y/∂y+∂z/∂z)dxdydz =3∫∫∫dxdydz=3V where V is the volume of the ellipsoid well known to be 4/3π(abc)^1/2. Therefore the triple integral is 4π(abc)^1/2 but since only the upper half is asked we get 2π(abc)^1/2. We must notice that the ellipsoid as it is given is NOT in the usual form with squares in the denominators so we get a result with square root of abc. I cant make it clearer!
