Finding the average value of a function! over a region! Completely related to this question:
Finding surface area of part of a plane that lies inside a cylinder???

Find the average value of the function $f(x, y, z) = x^2yz$ over $S$

How does one do that?? I would think it would be just taking an average over the area, but the area is $S$ so I don't know how to be that specific!! Please help!!!

For the two linked problems, here are two views of a graph of the cylinder $ \ x^2 \ + \ y^2 \ = \ 3 \ $ and the oblique plane $ x \ + \ 2y \ + \ 3z \ = \ 1 \ $  :

In order to find the average value of $ \ f(x,y,z) \ = \ x^2 \ y \ z \ $ over the region $ \ S \ $ of the plane lying within the cylinder, it will help to "evaluate" the function on the plane.  We can re-write the equation of the plane to express $ \ z \ $ as the function $ \ z \ = \ \frac{1}{3} \ ( \ 1 \ - \ x \ - \ 2y \ ) \ $ .  We then want to integrate $ \ f \ $ over the projected area of $ \ S \ $ onto the $ \ xy-$ plane:
$$ \iint_S \ x^2 \ y \ \cdot \frac{1}{3}  ( \ 1 \ - \ x \ - \ 2y \ ) \ \ dA \ \ . $$
It will be convenient to use polar coordinates, since the projected area is simply a circle of radius $ \ \sqrt{3} \ $ centered on the origin; this gives us
$$ \frac{1}{3} \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^2 \ \cos^2  \theta) \ (r \ \sin \ \theta) \ ( \ 1 \ - \ r \ \cos \ \theta \ - \ 2r \ \sin \ \theta \ ) \ \ r \ dr \ d\theta   $$
$$ = \ \ \frac{1}{3} \ \left[ \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^4 \ \cos^2  \theta \ \sin \ \theta)  \ \  dr \ d\theta \ \ - \ \  \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^5 \ \cos^3  \theta \ \sin \ \theta) \ \  dr \ d\theta \quad - \ \ 2 \ \int_0^{2 \pi} \int_0^{\sqrt{3}} \ (r^5 \ \cos^2  \theta \ \sin^2  \theta)  \ \  dr \ d\theta \ \right] \ \ . $$
The first two integrals produce antiderivatives which are just powers of cosine, making the integral values zero over one full period.  So we only need to calculate
$$ -\frac{2}{3} \ \int_0^{2 \pi}  \cos^2  \theta \ \sin^2  \theta \ \ d\theta \  \int_0^{\sqrt{3}} \ r^5  \ \  dr \ = \ -\frac{2}{3} \ \cdot \ \frac{\pi}{4} \ \cdot \ \frac{27}{6} \ = \ -\frac{3  \pi}{4} \ \ . $$  
As for the average value of the function, I am taking it that since the question asks for the average to be taken over the region $ \ S \ $ , we would divide this last result by the area found by john in the linked problem, thus,
$$  \langle \ f \ \ \rangle_S \ = \ \frac{- \ 3 \pi / 4}{\pi \ \sqrt{14}} \ = \ -\frac{3}{4 \ \sqrt{14}} \  \ \text{or} \ \ -\frac{3 \ \sqrt{14}}{56} \ \ . $$
Is this a plausible result?  If we look at the function $ \ x^2 \ y \ z \ $ , its sign is determined by the factors $ \ y \ $ and $ \ z \ $ .  The oblique plane crosses the $ \ xy-$ plane on the line $ \ x \ + \ 2y \ = \ 1 \ $ , so $ \ z \ $ is negative "below" that line in the $ \ xy-$ plane and positive "above" it.  On the right below is a graph depicting the portions of the region $ \ S \ $ where $ \ x^2 \ y \ z \ $ is positive [green] and negative [red] ; as our region of integration is dominated by "negative" sections.  So it is credible that the average value $  \langle \ f \ \ \rangle_S \ $ should be negative.  The graph of $ \ \frac{1}{3} \ x^2 \ y \   ( \ 1 \ - \ x \ - \ 2y \ ) \ $ at left below supports this expectation.

