Question

Find the average value of f(x, y) = xy^2 over the rectangle R with vertices (0, −1), (2, −1), (2, 1) and (0, 1).​

Answer 1

The rectangle
`R` is the set

`R = \{(x,y) \mid 0 \le x \le 2 \text{ and } -1 \le y \le 1\}`

The average value of
`f(x,y)` over
`R` is the ratio of the integral of
`f` over
`R` to the area of
`R`.

`f_(\rm ave) = (\displaystyle \iint_R f(x,y) \, dA)/(\displaystyle \iint_R dA)`

The rectangle is 2-by-2, so its area is

`\displaystyle \iint_R dA = 2*2 = 4`

Integrate
`f`.

`\displaystyle \iint_R xy^2 \, dA = \int_(-1)^1 \int_0^2 xy^2 \, dx \, dy \\\\ ~~~~~~~~ = 2 \int_(-1)^1 y^2 \, dy \\\\ ~~~~~~~~ = 4 \int_0^1 y^2 \, dy = \frac43`

Then the average value of
`f` is

`f_(\rm ave) = \frac{\frac43}4 = \boxed{\frac13}`