22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

2. Average Value on a Surface

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 (on 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 \(3x+2y-z=5\) over the region \(0 \le x \le 3\) and \(1 \le y \le 4\) if the temperature at any point on the plate is \(T=2x^2y\), as shown in the plots.

Plane in \(\mathbb{R}^3\)
AvgVal_example2
 
Temperature on the Plane
AvgVal_example
PY: Check the 3D plot. Make y axis go down to 0. Make Temp constrained scaling.

Since the plane is given as a graph, we may parametrize it by \(\vec R(x,y)=\left\langle x,y,3x+2y-5\right\rangle\). 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: \[\begin{aligned} \iint_P &T\,dS =\iint_P T|\vec{N}|\,du\,dv \\ &=\int_1^4\int_0^3 2x^2y\sqrt{14}\,dx\,dy =\sqrt{14}\left[\dfrac{x^3}{3}\right]_{x=0}^3\left[y^2\dfrac{}{}\right]_{y=1}^4 \\ &=\sqrt{14}(9)(16-1)\,dy =135\sqrt{14} \end{aligned}\] Finally, we can find \(T_\text{ave}\): \[ T_\text{ave}=\dfrac{1}{A}\iint_P T\,dS =\dfrac{135\sqrt{14}}{9\sqrt{14}}=15 \]

Find the average value of the function \(f=\sqrt{4x^2+4y^2+1}\) on the piece of the elliptic paraboloid \(z=x^2+y^2\) below \(z=4\). The length of the normal vector and the area were found in previous exercises.

\(\displaystyle f_\text{ave}=\dfrac{216}{\left(17^{3/2}-1\right)} \approx3.13\)

The elliptic paraboloid is parametrized by: \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \] In a previous exercise the area was found to be: \[ A=\int_0^{2\pi}\int_0^2 r\sqrt{4r^2+1}\,dr\,d\theta =\dfrac{\pi}{6}\left(17^{3/2}-1\right) \approx36.2 \] We want to compute the average of the function \[ f=\sqrt{4x^2+4y^2+1}=\sqrt{4r^2+1} \] Its integral is: \[\begin{aligned} \iint_S f\,dS &=\int_0^{2\pi}\int_0^2 \sqrt{4r^2+1}\,r\sqrt{4r^2+1}\,dr\,d\theta \\ &=2\pi\int_0^2 (4r^3+r)\,dr =2\pi\left[r^4+\dfrac{r^2}{2}\right]_0^2 \\ &=2\pi\left(16+2\right) =36\pi \end{aligned}\] So the average value of \(f\) is: \[\begin{aligned} f_\text{ave}&=\dfrac{1}{A}\iint_S f\,dS =\dfrac{6}{\left(17^{3/2}-1\right)\pi}36\pi \\ &=\dfrac{216}{\left(17^{3/2}-1\right)} \approx3.13 \end{aligned}\]

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