Skip to main content

Section 1.4 Area of planar regions and average value

Our definitions of area and average value arise directly from looking at the Riemann sum definition of the integral.

Definition 1.4.1. Area of planar region.

A bounded region \(\mathcal{R}\subseteq \R^2\) is measurable if the constant function \(f(x,y)=1\) is integrable over \(\mathcal{R}\text{.}\) In this case we define the area of \(\mathcal{R}\) as

\begin{equation} \operatorname{area}\mathcal{R}=\iint\limits_\mathcal{R} 1 \, dA\text{.}\tag{1.4.1} \end{equation}

Remark 1.4.2.

We have seen other notions of area before in our calculus journey. Definition 1.4.1 is more general (it covers a wider family of regions \(\mathcal{R}\subseteq\R^2\)) and can be shown to agree to the previous definitions for the cases they cover.

Example 1.4.3. Area of planar regions.

Use Definition 1.4.1 to compute the area of the given region \(\mathcal{R}\subseteq \R^2\text{.}\)

  1. \(\mathcal{R}\) is the bounded region between the graphs of \(x=y^2\) and \(x=\frac{1}{2}y^2+1\text{.}\)

  2. \(\mathcal{R}\) is the disk \(x^2+y^2\leq R^2\) of radius \(R\) centered at the origin.

Solution.

  1. The region is type 2. (See solution to Example 1.3.8 for diagram and details.) We have

    \begin{align*} \operatorname{area}\mathcal{R} \amp =\iint\limits_\mathcal{R} 1\, dA \\ \amp = \int_{-\sqrt{2}}^{\sqrt{2}}\int\limits_{y^2}^{y^2/2+1}1\, dx\, dy \\ \amp = \int_{-\sqrt{2}}^{\sqrt{2}}1-\frac{1}{2}y^2\, dy\\ \amp = \frac{4\sqrt{2}}{3} \end{align*}

    .

  2. First we describe \(\mathcal{R}\) as a type-1 region:

    \begin{equation*} \mathcal{R}=\{(x,y)\colon -R\leq x\leq R, -\sqrt{R-x^2}\leq y\leq \sqrt{R-x^2}\}\text{.} \end{equation*}

    Now compute

    \begin{align*} \operatorname{area}\mathcal{R} \amp= \iint\limits_{\mathcal{R}}1\, dA \\ \amp=\int_{-R}^R\int_{-\sqrt{R-x^2}}^{\sqrt{R-x^2}}1\, dy\, dx \\ \amp = \int_{-R}^R2\sqrt{R-x^2}\, dx\\ \amp = \int_{-\pi/2}^{\pi/2} 2\sqrt{R^2-R^2\sin^2\theta}\, R\cos\theta\, d\theta \amp (x=R\sin\theta, dx=R\cos\theta\, d\theta)\\ \amp = 2R^2\int_{-\pi/2}^{\pi/2}\abs{\cos\theta}\cos\theta\, d\theta\\ \amp =2R^2\int_{-\pi/2}^{\pi/2}\cos^2\theta\, d\theta \amp (\theta\in [0,\pi/2]\implies \cos\theta\geq 0)\\ \amp =2R^2\int_{-\pi/2}^{\pi/2}\frac{1}{2](1+\cos 2\theta)}\, d\theta\\ \amp =\pi R^2\text{.} \end{align*}

    We've just derived the familiar area formula for a disk of radius \(R\text{!}\)

Definition 1.4.4. Average valueAverage value over planar region

Assume \(f\) is integrable over the measurable region \(\mathcal{R}\subseteq \R^2\text{.}\) The average value of \(f\) over \(\mathcal{R}\text{,}\) denoted \(\operatorname{avg}_\mathcal{R}(f)\text{,}\) is defined as

\begin{equation} \operatorname{avg}_\mathcal{R}(f)=\frac{1}{\operatorname{area} \mathcal{R}}\iint\limits_\mathcal{R}f(x,y)\, dA\text{.}\tag{1.4.2} \end{equation}

Example 1.4.5. Average temperature over region.

The temperature \(T\) (in degrees Celsius) at a point \((x,y)\) in the bounded region \(\mathcal{R}\) between the graphs of \(y=1-x^2\) and \(y=x^2-1\) is given by \(T=f(x,y)=x^2+1\text{.}\) Compute the average temperature over \(\mathcal{R}\text{.}\)

Solution.

Observe that the curves \(y=1-x^2\) and \(y=x^2-1\) intersect at the points \((1,0)\) and \((-1,0)\text{,}\) and that we have \(x^2-1\leq 1-x^2\) for \(x\in [-1,1]\text{.}\) We describe \(\mathcal{R}\) as a type-1 region:

\begin{equation*} \mathcal{R}=\{(x,y)\colon -1\leq x\leq 1, x^2-1\leq y\leq 1-x^2\text{.} \end{equation*}

First compute the area of \(\mathcal{R}\text{:}\)

\begin{align*} \operatorname{area}\mathcal{R} \amp=\iint\limits_{\mathcal{R}}1\, dA \\ \amp=\int_{-1}^1\int_{x^2-1}^{1-x^2}1\, dy\, dx \\ \amp = \int_{-1}^12(1-x^2)\, dx\\ \amp =\frac{8}{3}\text{.} \end{align*}

Now compute

\begin{align*} \operatorname{avg}_{\mathcal{R}} T \amp=\frac{1}{\operatorname{area}\mathcal{R}}\iint\limits_{\mathcal{R}} x^2+1\, dA \\ \amp=\frac{3}{8}\int_{-1}^1\int_{x^2-1}^{1-x^2}x^2+1\, dy\, dx \\ \amp = \frac{3}{8}\int_{-1}^12x^2(1-x^2)+2(1-x^2)\, dx\\ \amp = \frac{3}{4}\int_{-1}^1 2(1-x^2)(1+x^2)\, dx\\ \amp =\frac{3}{2}(x-\frac{1}{5}x^5)\Bigr\vert_{0}^1\\ \amp = \frac{6}{5} \text{ degrees celcius}\text{.} \end{align*}