A L T A I R

Areas and Volumes

Theorem 8.1.1.  The area $A$ of the region $S$ between the graph of the continuous function $f$ and the $x$-axis is

$\displaystyle A = \int_{a}^{b} \left|\ f(x)\ \right| \operatorname{d}\!x$

where $a \le x \le b$.

Theorem 8.1.2.  The area between the curves $y = f(x)$ and $y = g(x)$ and between $x = a$ and $x = b$, where $a \le x \le b$, is

$\displaystyle A = \int_{a}^{b} \left|\ f(x) - g(x)\ \right| \operatorname{d}\!x$

Figure 1

Proof.  Consider the region $S$ that lies between two curves $y = f(x)$ and $y = g(x)$ and between the vertical lines $x = a$ and $x = b$, where $f(x) \ge g(x)$ for all $x$ in $\left[a,\ b\right]$.

Just as we did in Definition 7.2.1, we divide $S$ into $n$ strips of equal width and then we approximate the $i$th strip by a rectangle with base $\Delta x$ and height $f(x_{i}^{*}) - g(x_{i}^{*})$. The Riemann sum

$\displaystyle \sum_{i = 1}^{n}\left[ f(x_{i}^{*}) - g(x_{i}^{*}) \right] \Delta x$

is therefore an approximation of the area of $S$.

This approximation appears to become better and better as $n \to \infty$. Therefore the area of the region $S$ is the limiting value of the sum of the areas of these approximating rectangles.

$\displaystyle A = \lim_{n \to \infty}\sum_{i = 1}^{n}\left[ f(x_{i}^{*}) - g(x_{i}^{*}) \right] \Delta x$

And by Definition 7.2.1,

$\displaystyle \begin{align} A &= \lim_{n \to \infty}\sum_{i = 1}^{n}\left[ f(x_{i}^{*}) - g(x_{i}^{*}) \right] \Delta x \\[7pt] &= \int_{a}^{b}\left[f(x) - g(x)\right] \operatorname{d}\!x \qquad \cdots \qquad \text{(i)} \end{align}$

If $g(x) \ge f(x)$, since heights must be positive, the height of the rectangles is $g(x_{i}^{*}) - f(x_{i}^{*})$. Therefore

$\displaystyle \begin{align} A &= \lim_{n \to \infty}\sum_{i = 1}^{n}\left[ g(x_{i}^{*}) - f(x_{i}^{*}) \right] \Delta x \\[7pt] &= \int_{a}^{b}\left[g(x) - f(x)\right] \operatorname{d}\!x \qquad \cdots \qquad \text{(ii)} \end{align}$

By $\text{(i)}$ and $\text{(ii)}$, the area $A$ is

$\displaystyle A = \int_{a}^{b}\left|\ f(x) - g(x)\ \right| \operatorname{d}\!x$

$\Box$

Example 8.1.3.  Find the area of the region bounded by the curves $y = \sin x$, $y = \cos x$, $x = 0$, and $x = \pi/2$.

Solution.  The points of intersection occur when $\sin x = \cos x$, that is, when $x = \pi/4$ (since $0 \le x \le \pi/2$). Notice that $\cos x \ge \sin x$ when $0 \le x \le \pi/4$ but $\sin x \ge \cos x$ when $\pi/4 \le x \le \pi/2$. Therefore the required area is

$\displaystyle \begin{align} A &= \int_{0}^{\large \frac{\pi}{2}} \left|\ \cos x - \sin x\ \right| \operatorname{d}\!x \\[7pt] &= \int_{0}^{\large \frac{\pi}{4}}(\cos x - \sin x) \operatorname{d}\!x + \int_{\large \frac{\pi}{4}}^{\large \frac{\pi}{2}}(\sin x - \cos x) \operatorname{d}\!x \\[7pt] &= \Big[ \sin x + \cos x \Big]_{0}^{\pi/4} + \Big[ -\cos x - \sin x \Big]_{\pi/4}^{\pi/2} \\[7pt] &= \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 0 - 1 \right) + \left( - 0 - 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \\[7pt] &= 2\sqrt{2} - 2 \end{align}$

$\Box$

Theorem 8.1.4.  Let $S$ be a solid that lies between $x = a$ and $x = b$ ($a \le x \le b$). If the cross-sectional area of $S$ in the plane $P_{x}$, through $x$ and perpendicular to the $x$-axis, is $A(x)$, where $A$ is a continuous function, then the volume of $S$ is

$\displaystyle V = \lim_{n \to \infty} \sum_{i = 1}^{n} A(x_{i}^{*}) \Delta x = \int_{a}^{b}A(x) \operatorname{d}\!x$

Figure 2

Figure 3

Theorem 8.1.5.  The volume of the solid, obtained by rotating about the $x$-axis the region between the curve $y = f(x)$ and the $x$-axis from $a$ to $b$, is

$\displaystyle V = \int_{a}^{b}\pi\left[f(x)\right]^{2} \operatorname{d}\!x$

Also, the volume of the solid obtained by rotating about the $y$-axis the region between the curve $y = f(x)$ and the $x$-axis from $a$ to $b$, is

$\displaystyle V = \int_{a}^{b}2\pi \left|\ x\ \right|f(x) \operatorname{d}\!x$

The body text and the figures were excerpted from James Stewart, Calculus 8E.