Area Between Curves

The area between the graphs of two functions is equal to the integral of a function, [latex]f(x)[/latex], minus the integral of the other function, [latex]g(x)[/latex]: [latex]A = \int_a^{b} ( f(x) - g(x) ) \, dx[/latex]. Area is a quantity that expresses the extent of a two-dimensional surface or shape, or planar lamina, in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).

Area Between Curves

The area between a positive-valued curve and the horizontal axis, measured between two values [latex]a[/latex] and [latex]b[/latex] ([latex]b[/latex] is defined as the larger of the two values) on the horizontal axis, is given by the integral from [latex]a[/latex] to [latex]b[/latex] of the function that represents the curve. The area between the graphs of two functions is equal to the integral of one function, [latex]f(x)[/latex], minus the integral of the other function, [latex]g(x)[/latex]:[latex]A=\int_a^{b}(f(x)−g(x))\,dx[/latex] where [latex]f(x)[/latex] is the curve with the greater y-value.

Example

Find the area between the two curves [latex]f(x)=x[/latex] and [latex]f(x)= 0.5 \cdot x^2[/latex] over the interval from [latex]x=0[/latex] to [latex]x=2[/latex]. Two curves, [latex]y=x[/latex] and [latex]y = 0.5 \cdot x^2[/latex], meet at the points [latex](x_0,y_0)=(0,0) [/latex] and [latex](x_1,y_1)=(2,2)[/latex]. Since [latex]x > 0.5 \cdot x^2[/latex] over the interval from [latex]x=0[/latex] to [latex]x=2[/latex], the area can be calculated as follows: [latex-display]\displaystyle{A = \int_0^{2} ( x - \frac{1}{2} x^2 ) \, dx = \left [ \frac{1}{2} x^2- \frac{1}{6} x^3 \right ]_{x=0}^{x=2} = \frac{2}{3}}[/latex-display]

Volumes

Volumes of complicated shapes can be calculated using integral calculus if a formula exists for the shape's boundary. Volume is the quantity of three-dimensional space enclosed by some closed boundary—for example, the space that a substance or shape occupies or contains. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. The volumes of more complicated shapes can be calculated using integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in three-dimensional space. A volume integral is a triple integral of the constant function [latex]1[/latex], which gives the volume of the region [latex]D[/latex]. That is to say: [latex-display]\displaystyle{\text{volume}(D)=\int\int\int\limits_D dx\,dy\,dz}[/latex-display] It can also mean a triple integral within a region [latex]D[/latex] in [latex]R^3[/latex] of a function [latex]f(x,y,z)[/latex], and is usually written as: [latex-display]\displaystyle{\iiint\limits_D f(x,y,z)\,dx\,dy\,dz}[/latex-display]

Example

The volume of the cuboid with side lengths 4, 5, and 6 may be obtained in either of two ways.

Method 1

Using the triple integral given above, the volume is equal to: [latex-display]\displaystyle{\iiint_\mathrm{cuboid} 1 \, dx\, dy\, dz}[/latex-display] of the constant function [latex]1[/latex] calculated on the cuboid itself. This yields: [latex-display]\displaystyle{\int_{z=0}^{z=5} \int_{y=0}^{z=6} \int_{x=0}^{x=4} 1 \, dx\, dy\, dz = 120}[/latex-display]

Method 2

Alternatively, we can use the double integral: [latex-display]\displaystyle{\iint_D 5 \ dx\, dy}[/latex-display] of the function [latex]z = f(x, y) = 5[/latex] calculated in the region [latex]D[/latex] in the [latex]xy[/latex]-plane, which is the base of the cuboid. For example, if a rectangular base of such a cuboid is given via the [latex]xy[/latex] inequalities [latex]3 \leq x \leq 7[/latex], [latex]4 \leq y \leq 10[/latex], our above double integral now reads: [latex-display]\displaystyle{\int_4^{10}\left( \int_3^7 \ 5 \ dx\right) dy =120}[/latex-display] This is the volume under the surface.

Average Value of a Function

The average of a function [latex]f(x)[/latex] over an interval [latex][a,b][/latex] is [latex]\bar f = \frac{1}{b-a} \int_a^b f(x) \ dx[/latex]. An average is a measure of the "middle" or "typical" value of a data set. It is a measure of central tendency. In the most common case, the data set is a discrete set of numbers. The average of a list of numbers is a single number intended to typify the numbers in the list, which is called the arithmetic mean. However, the concept of average can be extended to functions as well. If [latex]n[/latex] numbers are given, each number denoted by [latex]a_i[/latex], where [latex]i = 1, \cdots, n[/latex], the arithmetic mean is the sum of all [latex]a_i[/latex] values divided by [latex]n[/latex]: [latex-display]\displaystyle{AM=\frac{1}{n}\sum_{i=1}^na_i}[/latex-display] extend this definition to continuum by making the following substitution: [latex-display]\displaystyle{\sum \rightarrow \int, a_i \rightarrow f(x), \frac{1}{n} \rightarrow \frac{dx}{b-a}}[/latex-display] Therefore, the average of a function [latex]f(x)[/latex] over an interval [latex][a,b][/latex] (where [latex]b > a[/latex]) is expressed as: [latex-display]\displaystyle{\bar f = \frac{1}{b-a} \int_a^b f(x) \ dx}[/latex-display] Note that the average is equal to the area under the curve, [latex]S[/latex], divided by the range: [latex-display]\displaystyle{\frac{S}{b-a}}[/latex-display]

Mean Value Theorem for Integration

The first mean value theorem for integration states that if [latex]G: [a, b] \to R[/latex] is a continuous function and [latex]\varphi[/latex] is an integrable function that does not change sign on the interval [latex](a, b)[/latex], then there exists a number [latex]x[/latex] in [latex](a, b)[/latex] such that: [latex-display]\displaystyle{\int_a^b G(t)\varphi (t) \, dt=G(x) \int_a^b \varphi (t) \, dt}[/latex-display] In particular, if [latex]\varphi(t) = 1[/latex] for all [latex]t[/latex] in [latex][a, b] [/latex], then there exists [latex]x[/latex] in [latex](a, b)[/latex] such that: [latex-display]\displaystyle{\int_a^b G(t) \, dt=\ G(x)(b - a)}[/latex-display] The value [latex]G(x)[/latex] is the mean value of [latex]G(t)[/latex] on [latex][a, b] [/latex] as we saw previously.

Cylindrical Shells

In the shell method, a function is rotated around an axis and modeled by an infinite number of cylindrical shells, all infinitely thin. Shell integration (also called the shell method) is a means of calculating the volume of a solid of revolution when integrating perpendicular to the axis of revolution. (When integrating parallel to the axis of revolution, you should use the disk method. ) While less intuitive than disk integration, it usually produces simpler integrals. Intuitively speaking, part of the graph of a function is rotated around an axis, and is modeled by an infinite number of cylindrical shells, all infinitely thin. The idea is that a "representative rectangle" (used in the most basic forms of integration, such as [latex]\int x \,dx[/latex]) can be rotated about the axis of revolution, thus generating a hollow cylinder with infinitesimal volume. Integration, as an accumulative process, can then calculate the integrated volume of a "family" of shells (a shell being the outer edge of a hollow cylinder), giving us the total volume. The volume of the solid formed by rotating the area between the curves of [latex]f(x)[/latex] and [latex]g(x)[/latex] and the lines [latex]x=a[/latex] and [latex]x=b[/latex] about the [latex]y[/latex]-axis is given by: [latex-display]\displaystyle{V = 2\pi \int_a^b x \left | f(x) - g(x) \right | \,dx}[/latex-display] In the integrand, the factor [latex]x[/latex] represents the radius of the cylindrical shell under consideration, while  is equal to the height of the shell. Therefore, the entire integrand, [latex]2\pi x \left | f(x) - g(x) \right | \,dx[/latex], is nothing but the volume of the cylindrical shell. By adding the volumes of all these infinitely thin cylinders, we can calculate the volume of the solid formed by the revolution. The volume of solid formed by rotating the area between the curves of [latex]f(y)[/latex] and and the lines [latex]y=a[/latex] and [latex]y=b[/latex] about the [latex]x[/latex]-axis is given by: [latex-display]\displaystyle{V = 2\pi \int_a^b x \left | f(y) - g(y) \right | \,dy}[/latex-display]

Work

Forces may do work on a system. Work done by a force ([latex]F[/latex]) along a trajectory ([latex]C[/latex]) is given as [latex]\int_C \mathbf{F} \cdot d\mathbf{x}[/latex]. For moving objects, the rate of the work done by a force (measured in joules/second, or watts) is the scalar product of the force (a vector ) and the velocity vector of the point of application. This scalar product of force and velocity is classified as instantaneous power delivered by the force. Just as velocities may be integrated over time to obtain a total distance, by the fundamental theorem of calculus, the total work along a path is similarly the time-integral of instantaneous power applied along the trajectory of the point of application. Work is the result of a force on a point that moves through a distance. As the point moves, it follows a curve [latex]X[/latex], with a velocity [latex]v[/latex], at each instant. The small amount of work [latex]\delta W[/latex] that occurs over an instant of time [latex]\delta t[/latex] is calculated as: [latex-display]\delta W = \mathbf{F}\cdot\mathbf{v}\delta t[/latex-display] where the term [latex]\mathbf{F}\cdot\mathbf{v}[/latex] is the power over the instant [latex]\delta t[/latex]. The sum of these small amounts of work over the trajectory of the point yields the work: [latex-display]\displaystyle{W = \int_{t_1}^{t_2}\mathbf{F} \cdot \mathbf{v}dt}[/latex-display] [latex-display]\displaystyle{\quad = \int_{t_1}^{t_2}\mathbf{F} \cdot {\frac{d\mathbf{x}}{dt}}dt}[/latex-display] [latex-display]\displaystyle{\quad =\int_C \mathbf{F} \cdot d\mathbf{x}}[/latex-display] where [latex]C[/latex] is the trajectory from [latex]x(t_1)[/latex] to [latex]x(t_2)[/latex]. This integral is computed along the trajectory of the particle, and is therefore said to be path-dependent. If the force is always directed along this line, and the magnitude of the force is [latex]F[/latex], then this integral simplifies to: [latex-display]\displaystyle{W=\int_C \mathbf{F} \, ds}[/latex-display] where [latex]s[/latex] is distance along the line. If [latex]F[/latex] is constant, in addition to being directed along the line, then the integral simplifies further to: [latex-display]\displaystyle{W = \int_C Fds = F\int_C ds = Fd}[/latex-display] This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot product [latex]F·dx = F \cdot \cos \theta \cdot dx[/latex], where [latex]\theta[/latex] is the angle between the force vector and the direction of movement. This is to say: [latex-display]\displaystyle{W = \int_C \mathbf{F} \cdot d\mathbf{x} = Fd\cos\theta}[/latex-display]

Example: Work Done by a Spring

Let's consider an object with mass [latex]m[/latex] attached to an ideal spring with spring constant [latex]k[/latex]. When the object moves from [latex]x=x_0[/latex] to [latex]x=0[/latex], work done by the spring would be: [latex-display]\displaystyle{W = \int_C \mathbf{F_s} \cdot d\mathbf{x} = \int_{x_0}^{0} (-kx)dx = \frac{1}{2} k x_0^2}[/latex-display]

Volumes of Revolution

Disc and shell methods of integration can be used to find the volume of a solid produced by revolution. A solid of revolution is a solid figure obtained by rotating a plane curve around some straight line (the axis) that lies on the same plane. Here, we will study how to compute volumes of these objects. Two common methods for finding the volume of a solid of revolution are the disc method and the shell method of integration. To apply these methods, it is easiest to draw the graph in question; identify the area that is to be revolved about the axis of revolution; determine the volume of either a disc-shaped slice of the solid, with thickness [latex]\delta x[/latex], or a cylindrical shell of width [latex]\delta x[/latex]; and then find the limiting sum of these volumes as [latex]\delta x[/latex] approaches [latex]0[/latex], a value which may be found by evaluating a suitable integral.

Disc Method

The disc method is used when the slice that was drawn is perpendicular to the axis of revolution; i.e. when integrating parallel to the axis of revolution. The volume of the solid formed by rotating the area between the curves of [latex]f(x)[/latex] and [latex]g(x)[/latex] and the lines [latex]x=a[/latex] and [latex]x=b[/latex] about the [latex]x[/latex]-axis is given by: [latex]\displaystyle{V = \pi \int_a^b \left | f^2(x) - g^2(x) \right | \,dx}[/latex] If [latex]g(x) = 0[/latex] (e.g. revolving an area between curve and [latex]x[/latex]-axis), this reduces to: [latex-display]\displaystyle{V = \pi \int_a^b f(x)^2 \,dx}[/latex-display] The method can be visualized by considering a thin horizontal rectangle at [latex]y[/latex]between [latex]y=f(x)[/latex] on top and [latex]y=g(x)[/latex] on the bottom, and revolving it about the [latex]y[/latex]-axis; it forms a ring (or disc in the case that [latex]g(x)=0[/latex]), with outer radius [latex]f(x)[/latex] and inner radius [latex]g(x)[/latex]. The area of a ring is: [latex-display]\pi (R^2 - r^2)[/latex-display] where [latex]R[/latex] is the outer radius (in this case [latex]f(x)[/latex]), and [latex]r[/latex] is the inner radius (in this case [latex]g(x)[/latex]). Summing up all of the areas along the interval gives the total volume. Alternatively, where each disc has a radius of [latex]f(x)[/latex], the discs approach perfect cylinders as their height [latex]dx[/latex] approaches zero. The volume of each infinitesimal disc is therefore: [latex-display]\pi f^2(x) dx[/latex-display] An infinite sum of the discs between [latex]a[/latex] and [latex]b[/latex] manifests itself as the integral seen above, replicated here: [latex-display]\displaystyle{V = \pi \int_a^b f(x)^2 \,dx}[/latex-display]

Shell Method

The shell method is used when the slice that was drawn is parallel to the axis of revolution; i.e. when integrating perpendicular to the axis of revolution. The volume of the solid formed by rotating the area between the curves of [latex]f(x)[/latex]and [latex]g(x)[/latex] and the lines [latex]x=a[/latex] and [latex]x=b[/latex] about the [latex]y[/latex]-axis is given by: [latex-display]\displaystyle{V = 2\pi \int_a^b x \left | f(x) - g(x) \right | \,dx}[/latex-display] If [latex]g(x)=0[/latex] (e.g. revolving an area between curve and [latex]x[/latex]-axis), this reduces to: [latex-display]\displaystyle{V = 2\pi \int_a^b x \left | f(x) \right | \,dx}[/latex-display] The method can be visualized by considering a thin vertical rectangle at [latex]x[/latex] with height [latex][f(x)-g(x)][/latex] and revolving it about the [latex]y[/latex]-axis; it forms a cylindrical shell. The lateral surface area of a cylinder is [latex]2 \pi r h[/latex], where [latex]r[/latex] is the radius (in this case [latex]x[/latex]), and [latex]h[/latex] is the height (in this case [latex][f(x)-g(x)][/latex]). Summing up all of the surface areas along the interval gives the total volume.