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
$$=\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}(x\frac{\delta(y,z)}{\delta(\theta,\psi)}+y\frac{\delta(z,x)}{\delta(\theta,\psi)}+z\frac{\delta(x,y)}{\delta(\theta,\psi)})\,d\theta\, d\psi$$
$$=\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}(\sin \theta)\,d\theta\, d\psi$$
$$=-abc\pi$$
You have a mistake in $\frac{\delta(x,y)}{\delta(\theta,\psi)}=ab\sin\theta \cos\psi$.
It should be $\frac{\delta(x,y)}{\delta(\theta,\psi)}=ab\sin\theta \cos\theta$
Also you must write the order of integral correctly given $0 \leq \psi \leq 2\pi$ and $0 \leq \theta \leq \pi/2$.
This is the integral you should finally get:
$\displaystyle \int_0^{2\pi} \int_0^{\pi/2} abc \sin\theta \ d\theta \ d\psi = 2 \pi abc$