22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

4. Average Value

Just as we can find the average value of a function over a region in a plane or a region in space or on a curve, we can also find the average value of a function over a surface. For example, we may wish to find the average temperature of a plate or a bowl given the temperature function or find the average density of a surface if we are given its mass density function. In such cases, we first calculate the total value of the quantity by integrating the quantity over the region. Then we divide by the total surface area to find the amount of the quantity per unit area (or average). \[ f_\text{ave}=\dfrac{1}{A}\iint_S f\,dS =\dfrac{\displaystyle\iint_S f(\vec R(u,v))\,|\vec{N}|\,du\,dv} {\displaystyle\iint_S |\vec{N}|\,du\,dv} \]

For example, suppose we want to find the average temperature on a plate, \(P\), lying in the plane \(3x+2y-z=2\) over the region \(0 \le x \le 3\) and \(1 \le y \le 4\) if the temperature at any point on the plate is \(T=270+2x^2y\), as shown in the plots.

PY: Avek: Redo the 3D plot. the surface has changed. Try to make a single 3D plot with correct color function. Make y axis go down to 0. Use constrained scaling if possible.

Since the plane is given as the graph of \(z=3x+2y-2\), we may parametrize it by \(\vec R(x,y)=\left\langle x,y,3x+2y-2\right\rangle\). Then \[\begin{aligned} \vec{e}_x&=\left\langle 1,0,3\right\rangle \\ \vec{e}_y&=\left\langle 0,1,2\right\rangle \\ \vec{N}&=\vec{e}_x\times\vec{e}_y=\left\langle -3,-2,1\right\rangle \\ |\vec{N}|&=\sqrt{9+4+1}=\sqrt{14} \end{aligned}\] We can now find the surface area of the section of the plane we are interested in: \[\begin{aligned} A&=\iint_P \,dS =\iint_P |\vec{N}|\,du\,dv \\ &=\int_1^4\int_0^3 \sqrt{14}\,dx\,dy =\sqrt{14}\cdot3\cdot3=9\sqrt{14} \end{aligned}\] Now we need to integrate the temperature function over the region: \[\begin{aligned} \iint_P &T\,dS =\iint_P T|\vec{N}|\,du\,dv \\ &=\int_1^4\int_0^3 (270+2x^2y)\sqrt{14}\,dx\,dy \\ &=270(\text{Area}) +\sqrt{14}\left[\dfrac{x^3}{3}\right]_{x=0}^3\left[y^2\dfrac{}{}\right]_{y=1}^4 \\ &=270(9\sqrt{14})+\sqrt{14}(9)(16-1) \\ &=270(9\sqrt{14})+135\sqrt{14} \end{aligned}\] Finally, we can find \(T_\text{ave}\): \[\begin{aligned} T_\text{ave}&=\dfrac{1}{A}\iint_P T\,dS =\dfrac{270(9\sqrt{14})+135\sqrt{14}}{9\sqrt{14}} \\ &=270+15=285 \end{aligned}\]