Question

Find the average value of f(x,y)=5 x^2 y^4 over the rectangle R with vertices (−2,0),(−2,6),(2,0),(2,6).
Average value =

Answer 1

The rectangle $R$ with vertices $(-2,0),(-2,6),(2,0)$ and $(2,6)$ is as shown in the figure below:

The area of the rectangle is,

$A(R)=(2+2)(6-0)=(4)(6)=24$

The average value of function over the rectangle $R$ is evaluated as,

$f_{ave}=\frac{1}{A(R)}\iint_{R}f(x,y)dA$

$=\frac{1}{24}\int_{-2}^{2}\int_{0}^{6}5x^2y^4dydx$

$=\frac{5}{24}[\frac{x^3}{3}]_{-2}^{2}[\frac{y^5}{5}]_{0}^{6}$

$=\frac{5}{24}(\frac{8}{3}+\frac{8}{3})(\frac{6^5}{5})$

$=\frac{5}{24}(\frac{16}{3})(\frac{7776}{5})$

$=1728$

Therefore, the average value of the function over the rectangle is $f_{ave}=1728$ .