I have the following question:
Find a non-linear function $z = f(x, y)$ such that the average value of $z$ is 3 for $-1 \le x \le 5$ and $-4 \le y \le 4$.
My first thought was the following:
$$ \frac{1}{[(5)-(-1)][(4)-(-4)]} \iint_{R} f(x, y) dA = 3 $$ $$ = \frac{1}{48} \iint_{R} f(x, y) dA = 3 $$ $$ = \iint_{R} f(x, y) dA = 144 $$
Then, break up the integral into x- and y-components:
$$ = \int_{x = -1}^5 \int_{y = -4}^4 f(x, y) dy dx = 144 $$
Then, if $ \int_{y = -4}^4 f(x, y) dy $ was equal to $24$, $ \int_{x = -1}^5 \int_{y = -4}^4 f(x, y) dy dx $ would be equal to $ 24x \Big|_{-1}^5 = 144 $.
However, I have no idea how I would create a non-linear function $f(x, y)$ such that $ \int_{y = -4}^4 f(x, y) dy $ would be equal to $24$. Am I even on the right track?
Thank you.