<s>[INST] How to find the average value of $f(x,y)$ on $R=\{(x,y): -1\le x\le1,0\le y\le3\}$ Given $R=\{(x,y): -1\le x\le1,0\le y\le3\}$
For the function $f(x,y)$ evaluate $\int\int_R f(x,y)dA$ and find the average value of $f(x,y$ on $R$
$f(x,y)=x^3+xy^3-3y$
My Try:
$=\int_{-1}^{1}\int_0^3x^3+xy^3-3ydydx$
$=\int_{-1}^{1}(x^3y+\dfrac{xy^4}{4}-\dfrac{3y^2}{2})\bigg|_0^3dx$
$=\int_{-1}^{1}3x^3+\dfrac{81x}{4}-\dfrac{27}{2}dx$
$=\dfrac{3x^4}{4}+\dfrac{81x^2}{8}-\dfrac{27x}{2}\bigg|_{-1}^{1}$
$=-\dfrac{54}{2}$
$=-27$
Is my above answer correct?
How to find the average value of $f(x,y)$ on $R$?
 [/INST] So, not quite!
First, we need to know how exactly the average value of a two variable function $f(x,y)$ is computed. Typically, the following equation is utilized to compute the average value of a function $f(x,y)$ over a rectangle $R$:
$$f_{ave} = \frac{1}{\text{Area}(R)} \iint_R f(x,y) dxdy$$
Since we're given a rectangle $R$ with side lengths $2$ and $3$, we find that $\text{Area(R)}$ = 6.
So, our equation above becomes:
$$f_{ave} = \frac{1}{6} \iint_R f(x,y) dxdy$$
Then, we need only simply compute $\iint_R f(x,y) dxdy$ which you have done correctly:
$$\iint_R f(x,y) dxdy = \int_{-1}^{1} \int_{0}^{3} (x^3 + 3xy^3 -3y)dydx = -27$$
So, your answer will simply be:
$f_{ave} = \frac{1}{6} (-27) = -\frac{27}{6}$
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