Maths question concerning three dimensional integration

[Drawoh Tesremos]

A short while ago I was asked to help someone with a three dimensional integration problem.

Specifically, we have a fuction F(x,y) = 16 - x - y^2, and the requirement is,
(a) determine the volume under a defined area, and
(b) determine the surface area within the same limits.

I have no problem with (a), it's a double integral of the function F, and the two sets of limits of integration are relatively easy to work out.

However, with part (b) I just cannot remember what it is that you have to integrate. The limits of integration are ok; they are the same as in part (a).

Can anyone here help? I recall doing this at uni, some 45 years ago, but haven't been able to locate my old text books.

 

[Drawoh Tesremos]

For completeness, though not necessary to be able to answer the question I'm asking in the previous post, the defined area is the triangle contained by the three lines
y = 0
x = 2
x = y

 

[Drawoh Tesremos]

Thanks, both of you. You've shown two completely different methods, which is not unusual, as many maths problems can be approached in more than one way. I'll look further at both of your suggestions. Thanks.

 

[Drawoh Tesremos]

Thanks for the further contributions on this. I'm away from home at the moment, but just checked in to see if there was anything further. Before leaving home I passed on the sqrt(1 + (dF/dx)^2 + (dF/dy)^2) suggestion as it seemed more workable than the polar co-ordinate idea. This particular question didn't seem to lend itself to polar co-ordinates due to the region to be integrated being a triangle. Had it been a circle, or similar, I may well have gone with the polar co-cordinates idea.

Thanks for the further comments. I'll look at them when I get back home.