Multiple integrals: Applications of multiple integrals
Average value of a function of 2 or 3 variables
The average value of an integrable function of two variables over a region \(R\) can be calculated as follows: \[\begin{aligned}\text{average value of }f\text{ over }R &=\frac{1}{\text{area of }R}\iint_R f(x,y)\,\dd(x,y)\\[0.25cm] &=\dfrac{\iint_R f(x,y)\,\dd(x,y)}{\iint_R \dd(x,y)}\end{aligned}\]
A very large number of points \((x,y)\) are randomly generated in the triangle \(T\) with vertices \((0,0)\), \(1,0)\), and \((0,10\). What is the average value of \(x+y\) for the set of points?
The average value of \(x+y\) for the set of points is equal to the average value of the function over the triangle, which is equal to\[\begin{aligned}\frac{1}{1/2}\iint_T (x+y)\,\dd(x,y) &= 2\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1-x}(x+y)\,\dd y\right)\dd x \\[0.25cm] &= 2\int_{x=0}^{x=1}\biggl[x\,y+\frac{1}{2}y^2\biggr]_{y=0}^{y=1-x}\;\dd x\\[0.25cm] &=2\int_{x=0}^{x=1} \left(x\,(x-1) - (1-x)^2\right)\,\dd x\\[0.25cm] &=\int_{x=0}^{x=1} \left(1-x^2\right)\,\dd x\\[0.25cm] &=\biggl[x-\frac{1}{3}x^3\biggl]_{x=0}^{x=1}\\[0.25cm] &=\frac{2}{3}\end{aligned}\]
The average value of an integrable function of three variables over a region \(R\) can be calculated as follows: \[\begin{aligned}\text{average value of}f\text{ over }R &=\frac{1}{\text{volume of }R}\iiint_R f(x,y,z)\,\dd(x,y,z)\\[0.25cm] &=\dfrac{\iiint_R f(x,y,z)\,\dd(x,y,z)}{\iint_R \dd(x,y)}\end{aligned}\]
