Chapter 7 Applications of Integration 7.1 Area of a Region Between Two Curves 7.2 Volume: The Disk Method 7.3 Volume: The Shell Method 7.5 Work 7.6 Moments, Centers of Mass, and Centroids 7.7 Fluid Pressure and Fluid Force
Arc Length A rectifiable curve is one that has a finite arc length. A sufficient condition for the graph of a function f to be rectifiable between (a, f a ) and (b, f b ) is that f be continuous on [a, b]. Such a function is continuous differentiable on [a, b], and its graph on the interval [a, b] is a smooth curve.
Arc Length When a function y = f(x) is continuously differentiable on the interval [a, b], the arc length will be defined by the following way: For the partition a = x 0 < x 1 < x 2 < < x n = b the arc length can be approximated by using the sum of lengths of line segments: s = n i=1 n i=1 x i x i 1 2 + y i y i 1 2 1 + Δy i Δx i Using the limit we have n s = lim n i=1 2 1 + Δy i Δx i (Δx i ). 2 (Δx i ) = a b 1 + f x 2 dx
Definition of Arc Length Let the function y = f(x) represent a smooth curve on the interval [a, b]. The arc length of f between a and b is s = a b 1 + f x 2 dx. Similarly, for a smooth curve x = g(y), the arc length of g between c and d is s = c d 1 + g y 2 dy.
Example 1. Find the arc length from (x 1, y 1 ) to (x 2, y 2 ) on the graph of f x = mx + b.
Example 2. Find the arc length of the graph y = x3 6 + 1 2x on the interval 1 2, 2.
Example 3. Find the arc length of the graph y 1 3 = x 2 on the interval [0, 8].
Example 4. Find the arc length of the graph of y = ln(cos x) from x = 0 to x = π/4.
Example 5. An electric cable is hung between two towers that are 200 feet apart. The cable takes the shape of a catenary whose equation is y = 150 cosh x 150 = 75 ex/150 + e x/150 Find the arc length of the cable between the two towers.
Surface Area of Revolution When the graph of a continuous function is revolved about a line, the resulting surface is a surface of revolution. In computing the area of a surface of revolution the lateral surface area of the frustum of a right circular cone which has the following formula S = π r 1 + r 2 L = 2π 1 2 r 1 + r 2 L
Surface Area of Revolution Now consider a function f that has a continuous derivative on the interval [a, b]. Let Δ be a partition of [a, b], with subintervals of width Δx i. Then the line segment of length ΔL i = Δx i 2 + Δy i 2 generates a frustum of a cone.
Surface Area of Revolution Let r i be the average radius of this frustum. By Intermediate Value Theorem, a point d i exists in the ith subinterval such that r i = f(d i ). So the change of the surface area is ΔS i = 2πr i ܮΔ i = 2π f d i Δx i 2 + Δy i 2 = 2πf d i 1 + Δy i Δx i Now the surface area can be approximated by 2 Δx i n i=1 ΔS i = n i=1 2πf d i 1 + Δy i Δx i 2 Δx i.
By taking the limit n we have S = 2π a b f x 1 + f x 2 dx
Definition of the Area of a Surface of Revolution Let y = f(x) have a continuous derivative on the interval [a, b]. The area S of the surface of revolution formed by revolving the graph of f about a horizontal or vertical axis is S = 2π a b r x 1 + f x 2 dx where r(x) is the distance between the graph of f and the axis of revolution. If x = g(y) on the interval [c, d], then the surface area is S = 2π c d r y 1 + g y 2 dy where r(y) is the distance between the graph of g and the axis of revolution.
Example 6. Find the area of the surface formed by revolving the graph of f(x) = x 3 on the interval [0, 1] about the x-axis.
Example 7. Find the area of the surface formed by revolving the graph of f(x) = x 2 on the interval [0, 2] about the y-axis.
7.4 Work Work Done by a Constant Force If an object is moved a distance D in the direction of an applied constant force F, then the work W done by the force is defined as W = FD. There are four fundamental types of forces gravitational, electromagnetic, strong nuclear, and weak nuclear. A force can be thought of as a push or a pull; a force changes the state of rest or state of motion of a body. For gravitational forces on Earth, it is common to use units of measure corresponding to the weight of an object.
7.4 Work Example 1. Determine the work done in lifting a 50-pound object 4 feet. System of Measurement Measure of Work Measure of Force Measure of Distance U.S. foot-pound pound (l) foot (ft) International joule (J) newton (N) meter (m) C-G-S erg Dyne (dyn) centimeter (cm Conversions: 1 ft-lb 1.35582 J = 1.35582 10 7 ergs 1 N = 10 5 dyn 0.22481 lb 1 J = 10 7 ergs 0.73756 ft-lb 1 lb 4.44822 N
7.4 Work Work Done by a Variable Force If an object is moved along a straight line by a continuously varying force F(x), then the work W done by the force as the object is moved from x = a to x = b is given by W = a b F(x) dx. Idea: Let Δ be a partition of the interval: a = x 0 < x 1 < x 2 < < x n 1 < x n = b. We assume the force is constant as F c i (x i 1 c i x i ) for each interval. Then ΔW i = F c i Δx i So, the total work is approximated by Thus, W n i=1 n W = lim n i=1 W i = n i=1 F c i Δx i = F c i Δx i a b F(x) dx.
7.4 Work Some Laws in Physics 1. Hooke s Law: The force F required to compress or stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length. That is, F = kd. Where the constant of proportionality k (the spring constant) depends on the specific nature of the spring.
7.4 Work Some Laws in Physics 2. Newton s Law of Universal Gravitation: The force F of attraction between two particles of masses m 1 and m 2 is proportional to the product of the masses and inversely proportional to the square of the distance d between the two particles. That is, F = G m 1m 2 d 2. When m 1 and m 2 are in kilograms and d in meters, F will be in newtons for a value of G = 6.67 10 11 cubic meter per kilogram-second squared, where G is the gravitational constant.
7.4 Work Some Laws in Physics 3. Coulomb s Law: The force F between two charges q 1 and q 2 in a vacuum is proportional to the product of the charges and inversely proportional to the square of the distance d between the two charges. That is, F = k q 1q 2 d 2. When q 1 and q 2 are given in electrostatic units and d in centimeters, F will be in dynes for a value of k = 1.
7.4 Work Some Laws in Physics 1. Hooke s Law: The force F required to compress or stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length. That is, F = kd. Where the constant of proportionality k (the spring constant) depends on the specific nature of the spring. Example 2. A force of 30 newtons compresses a spring 0.3 meter from its natural length of 1.5 meters. Find the work done in compressing the spring and additional 0.3 meter.
7.4 Work Example 4. A spherical tank of radius 8 feet is half full of oil that weights 50 pounds per cubic foot. Find the work required to pump all of the oil out through a hole in the top of the tank.
7.4 Work Example 5. A 20-foot chain weighting 5 pounds per foot is lying coiled on the ground. How much work is required to raise one end of the chain to a height of 20 feet so that it is fully extended.