Applications of Integrals

Area between two curves

Suppose we are given two functions $y_{1}=f(x)$ and $y_{2}=g(x)$ and we want to find the area between them on the interval $[a,b]$ . Also assume that $f(x)\geq g(x)$ for all $x$ on the interval $[a,b]$ . Begin by partitioning the interval $[a,b]$ into $n$ equal subintervals each having a length of $\Delta x={\frac {b-a}{n}}$ . Next choose any point in each subinterval, $x_{i}^{*}$ . Now we can 'create' rectangles on each interval. At the point $x_{i}*$ , the height of each rectangle is $f(x_{i}^{*})-g(x_{i}^{*})$ and the width is $\Delta x$ . Thus the area of each rectangle is ${\bigl [}f(x_{i}^{*})-g(x_{i}^{*}){\bigr ]}\Delta x$ . An approximation of the area, $A$ , between the two curves is

$A:=\sum _{i=1}^{n}{\Big [}f(x_{i}^{*})-g(x_{i}^{*}){\Big ]}\Delta x$ .

Now we take the limit as $n$ approaches infinity and get

$A=\lim _{n\to \infty }\sum _{i=1}^{n}{\Big [}f(x_{i}^{*})-g(x_{i}^{*}){\Big ]}\Delta x$

which gives the exact area. Recalling the definition of the definite integral we notice that

$A=\int \limits _{a}^{b}{\bigl (}f(x)-g(x){\bigr )}dx$ .

This formula of finding the area between two curves is sometimes known as applying integration with respect to the x-axis since the rectangles used to approximate the area have their bases lying parallel to the x-axis. It will be most useful when the two functions are of the form $y_{1}=f(x)$ and $y_{2}=g(x)$ . Sometimes however, one may find it simpler to integrate with respect to the y-axis. This occurs when integrating with respect to the x-axis would result in more than one integral to be evaluated. These functions take the form $x_{1}=f(y)$ and $x_{2}=g(y)$ on the interval $[c,d]$ . Note that $[c,d]$ are values of $y$ . The derivation of this case is completely identical. Similar to before, we will assume that $f(y)\geq g(y)$ for all $y$ on $[c,d]$ . Now, as before we can divide the interval into $n$ subintervals and create rectangles to approximate the area between $f(y)$ and $g(y)$ . It may be useful to picture each rectangle having their 'width', $\Delta y$ , parallel to the y-axis and 'height', $f(y_{i}^{*})-g(y_{i}^{*})$ at the point $y_{i}^{*}$, parallel to the x-axis. Following from the work above we may reason that an approximation of the area, $A$ , between the two curves is

$A:=\sum _{i=1}^{n}{\Big [}f(y_{i}^{*})-g(y_{i}^{*}){\Big ]}\Delta y$ .

As before, we take the limit as $n$ approaches infinity to arrive at

$A=\lim _{n\to \infty }\sum _{i=1}^{n}{\Big [}f(y_{i}^{*})-g(y_{i}^{*}){\Big ]}\Delta y$ ,

which is nothing more than a definite integral, so

$A=\int \limits _{c}^{d}{\bigl (}f(y)-g(y){\bigr )}dy$ .

Regardless of the form of the functions, we basically use the same formula.

Volume

If the function $A(x)$ is continuous on $[a,b]$ , then the volume $V_{S}$ of the solid $S$ is given by:

$V_{S}=\int \limits _{a}^{b}A(x)dx$

Example 1: A right cylinder

Figure 1

Now we will calculate the volume of a right cylinder using our new ideas about how to calculate volume. Since we already know the formula for the volume of a cylinder this will give us a "sanity check" that our formulas make sense. First, we choose a dimension along which to integrate. In this case, it will greatly simplify the calculations to integrate along the height of the cylinder, so this is the direction we will choose. Thus we will call the vertical direction $x$ (see Figure 1). Now we find the function, $A(x)$ , which will describe the cross-sectional area of our cylinder at a height of $x$ . The cross-sectional area of a cylinder is simply a circle. Now simply recall that the area of a circle is $\pi r^{2}$ , and so $A(x)=\pi r^{2}$ . Before performing the computation, we must choose our bounds of integration. In this case, we simply define $x=0$ to be the base of the cylinder, and so we will integrate from $x=0$ to $x=h$ , where $h$ is the height of the cylinder. Finally, we integrate:

${\begin{aligned}V_{\mathrm {cylinder} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{0}^{h}\pi r^{2}dx\\&=\pi r^{2}\int \limits _{0}^{h}dx\\&=\pi r^{2}x{\bigg |}_{x=0}^{h}\\&=\pi r^{2}(h-0)\\&=\pi r^{2}h\end{aligned}}$

This is exactly the familiar formula for the volume of a cylinder.

Example 2: A right circular cone

Figure 2: The cross-section of a right circular cone by a plane perpendicular to the axis of the cone is a circle.

For our next example we will look at an example where the cross sectional area is not constant. Consider a right circular cone. Once again the cross sections are simply circles. But now the radius varies from the base of the cone to the tip. Once again we choose $x$ to be the vertical direction, with the base at $x=0$ and the tip at $x=h$ , and we will let $R$ denote the radius of the base. While we know the cross sections are just circles we cannot calculate the area of the cross sections unless we find some way to determine the radius of the circle at height $x$ .

Figure 3: Cross-section of the right circular cone by a plane perpendicular to the base and passing through the tip.

Luckily in this case it is possible to use some of what we know from geometry. We can imagine cutting the cone perpendicular to the base through some diameter of the circle all the way to the tip of the cone. If we then look at the flat side we just created, we will see simply a triangle, whose geometry we understand well. The right triangle from the tip to the base at height $x$ is similar to the right triangle from the tip to the base at height $h$ . This tells us that ${\frac {r}{h-x}}={\frac {R}{h}}$ . So that we see that the radius of the circle at height $x$ is $r(x)={\frac {R}{h}}(h-x)$ . Now using the familiar formula for the area of a circle we see that $A(x)=\pi {\frac {R^{2}}{h^{2}}}(h-x)^{2}$ .

Now we are ready to integrate.

${\begin{aligned}V_{\mathrm {cone} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{0}^{h}\pi {\frac {R^{2}}{h^{2}}}(h-x)^{2}dx\\&=\pi {\frac {R^{2}}{h^{2}}}\int \limits _{0}^{h}(h-x)^{2}dx\end{aligned}}$

By u-substitution we may let $u=h-x$ , then $du=-dx$ and our integral becomes

${\begin{aligned}&&=\pi {\frac {R^{2}}{h^{2}}}\left(-\int \limits _{h}^{0}u^{2}du\right)\\&&=\pi {\frac {R^{2}}{h^{2}}}\left(-{\frac {u^{3}}{3}}{\bigg |}_{h}^{0}\right)\\&&=\pi {\frac {R^{2}}{h^{2}}}\left(-0+{\frac {h^{3}}{3}}\right)\\&&={\frac {\pi }{3}}R^{2}h\end{aligned}}$

Example 3: A sphere

Figure 4: Determining the radius of the cross-section of the sphere at a distance $|x|$ from the sphere's center.

In a similar fashion, we can use our definition to prove the well known formula for the volume of a sphere. First, we must find our cross-sectional area function, $A(x)$ . Consider a sphere of radius $R$ which is centered at the origin in $\mathbb {R} ^{3}$ . If we again integrate vertically then $x$ will vary from $-R$ to $R$ . In order to find the area of a particular cross section it helps to draw a right triangle whose points lie at the center of the sphere, the center of the circular cross section, and at a point along the circumference of the cross section. As shown in the diagram the side lengths of this triangle will be $R$ , $|x|$ , and $r$ . Where $r$ is the radius of the circular cross section. Then by the Pythagorean theorem $r={\sqrt {R^{2}-|x|^{2}}}$ and find that $A(x)=\pi (R^{2}-|x|^{2})$ . It is slightly helpful to notice that $|x|^{2}=x^{2}$ so we do not need to keep the absolute value.

So we have that

${\begin{aligned}V_{\mathrm {sphere} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{-R}^{R}\pi (R^{2}-x^{2})dx\\&=\pi \int \limits _{-R}^{R}R^{2}dx-\pi \int \limits _{-R}^{R}x^{2}dx\\&=\pi R^{2}x{\Bigg |}_{-R}^{R}-\pi {\frac {x^{3}}{3}}{\Bigg |}_{-R}^{R}\\&=\pi R^{2}(R-(-R))-\pi \left({\frac {R^{3}}{3}}-{\frac {(-R)^{3}}{3}}\right)\\&=2\pi R^{3}-{\frac {2\pi }{3}}R^{3}={\frac {4\pi }{3}}R^{3}\end{aligned}}$

Arc length

Suppose that $f'$ is continuous on $[a,b]$ . Then the length of the curve given by $y=f(x)$ between $a$ and $b$ is given by

$L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx$

And in Leibniz notation

$L=\int \limits _{a}^{b}{\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}dx$

Proof: Consider $y_{i+1}-y_{i}=f(x_{i+1})-f(x_{i})$ . By the Mean Value Theorem there is a point $z_{i}$ in $(x_{i+1},x_{i})$ such that

$y_{i+1}-y_{i}=f(x_{i+1})-f(x_{i})=f'(z_{i})(x_{i+1}-x_{i})$

So

${\bigl |}P_{i}P_{i+1}{\bigr |}$	$={\sqrt {(x_{i+1}-x_{i})^{2}+(y_{i+1}-y_{i})^{2}}}$
	$={\sqrt {(x_{i+1}-x_{i})^{2}+f'(z_{i})^{2}(x_{i+1}-x_{i})^{2}}}$
	$={\sqrt {{\bigl (}1+f'(z_{i})^{2}{\bigr )}(x_{i+1}-x_{i})^{2}}}$
	$={\sqrt {1+f'(z_{i})^{2}}}\Delta x$

Putting this into the definition of the length of $C$ gives

$L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\sqrt {1+f'(z_{i})^{2}}}\Delta x$

Now this is the definition of the integral of the function $g(x)={\sqrt {1+f'(x)^{2}}}$ between $a$ and $b$ (notice that $g$ is continuous because we are assuming that $f'$ is continuous). Hence

$L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx$

as claimed.

Example

Length of the curve $y=2x$ from $x=0$ to $x=1$}} As a sanity check of our formula, let's calculate the length of the "curve" $y=2x$ from $x=0$ to $x=1$ . First let's find the answer using the Pythagorean Theorem.

$P_{0}=(0,0)$

and

$P_{1}=(1,2)$

so the length of the curve, $s$ , is

$s={\sqrt {2^{2}+1^{2}}}={\sqrt {5}}$

Now let's use the formula

$s=\int \limits _{0}^{1}{\sqrt {1+\left({\tfrac {d(2x)}{dx}}\right)^{2}}}\,dx=\int \limits _{0}^{1}{\sqrt {1+2^{2}}}\,dx={\sqrt {5}}x{\bigg |}_{0}^{1}={\sqrt {5}}$

Resources

• Business Calculus: Application of definite integral, Rajendra Dahal (YouTube)
• Applications of Integrals - Cal I, Paul's Online Notes
• Applications of Integrals - Cal II, Paul's Online Notes

Licensing

Content obtained and/or adapted from:

• Area, Wikibooks: Calculus under a CC BY-SA license
• Volume, Wikibooks: Calculus under a CC BY-SA license
• Arc length, Wikibooks: Calculus under a CC BY-SA license