(a) Find the surface integral of $f=|x|-|y|$ over the part of $z=1-\frac{x^2}{M}-\frac{y^2}{N}$ inside the region $\frac{x^2}{M^2}+\frac{y^2}{N^2}=1$
(b) Find the surface integral of $f=|xy|$ over the part of $z=1-\frac{x^2}{M}-\frac{y^2}{N}$ inside the region $\frac{x^2}{M^2}+\frac{y^2}{N^2}=1$
I am assuming that if I can solve (a) I will be able to use a similar method to solve (b). So for (a), I have tried a change of variables using
$x=Mr$cos$t$, $y=Nr$sin$t$ and $z=1-Mr^2$cos$^2t-Nr^2$sin$^2t$.
Then I found the Jacobian to be $J=MNr$. So $dxdy=MNrdrdt$.
So then, the surface area element $=MNr\sqrt{1+4r^2}drdt$
Then, to avoid the absolute value functions, I split $f$ into a piecewise formula:
for $0\leq t\leq \frac{\pi}{2}$, $f=Mr$cos$t-Nr$sin$t$
for $\frac{\pi}{2}\leq t\leq \pi$, $f=-Mr$cos$t-Nr$sin$t$
for $\pi\leq t\leq \frac{3\pi}{2}$, $f=-Mr$cos$t+Nr$sin$t$
for $\frac{3\pi}{2}0\leq t\leq 2\pi$, $f=Mr$cos$t+Nr$sin$t$
and finally I tried to evaluate
DESIRED INTEGRAL $=\int_0^\frac{\pi}{2} \int_0^1 (Mr$cos$t-Nr$sin$t)MNr\sqrt{1+4r^2}drdt+$
$+ \int_\frac{\pi}{2}^\pi \int_0^1 (-Mr$cos$t-Nr$sin$t)MNr\sqrt{1+4r^2}drdt+$
$+ \int_\pi^\frac{3\pi}{2} \int_0^1 (-Mr$cos$t+Nr$sin$t)MNr\sqrt{1+4r^2}drdt+$
$+ \int_\frac{3\pi}{2}^2\pi \int_0^1 (Mr$cos$t+Nr$sin$t)MNr\sqrt{1+4r^2}drdt $
which can be done and I have no trouble doing. Does this method work at all? Have I made errors? Or do you have a better idea?