# Solving Problems using Line Integrals

Lesson overview

In the previous lesson, we defined the concept of a line integral and derived a formula for calculating them. We learned that line integrals give the volume between a surface $$f(x,y)$$ and a curve $$C$$. In this lesson, we'll learn about some of the applications of line integral for finding the volumes of solids and calculating work. In particular, we'll use the concept of line integrals to calculate the volume of a cylinder, the work done by a proton on another proton moving in the presence of its electric field, and the work done by gravity on a swinging pendulum.

Problem 1: Using a line integral, find the surface area of a cylinder (with its two end pieces missing) or tube with a height of$$h=2$$ and a radius of$$r=1$$.

Figure 1: A cylinder with height $$h=2$$ and radius $$r=1$$. The "top" and "bottom" pieces of the cylinder are removed.

The problem is asking us to find the surface area of a tube with a height of $$2$$ and a radius of $$1$$ using only the concept of a line integrl. Recall in the previous lesson that, basically, all a line integral ($$\int_Cf(s)dS$$) is is the area of a curved wall. The "sides" of the wall are created by taking a curve $$C$$ on the $$xy$$-plane and tracing it "up" in the vertical direction; the "base" of the wall is simply just the curve $$C$$; and the "ceiling" of the wall is the function $$f(s)$$. Since the shape of the cross-section of a tube is a circle and since the radius of our tube in this problem is just $$1$$, we know that the curve $$C$$ should just be a circle of radius $$1$$ on the $$xy$$-plane as illustrated in Figure 1. You could imagine that by moving the curve $$C$$ (in this case, a circle) in the upwards direction along the $$z$$-axis, this traces out the "sides" of the tube. Since we want the "ceiling to be circular, flat, and $$2$$ units above the $$xy$$-plane, let's graph the surface $$f(x,y)=2$$; the circular curve $$f(s)=2$$ will give the ceiling of the tube.

In general, we must use the expression

$$\int_{t_0}^tf(x(t),y(t))\sqrt{(x'(t))^2+(y'(t))^2}dt$$

to calculate a line integral; but since this example is so simple, we can just use the expression $$\int_Cf(s)dS$$ to calculate the line integral. If we substitute $$f(s)=2$$ into the equation above, we'll get

$$\int_Cf(s)dS=\int_C2dS=2\int_cdS.\tag{1}$$

The infinite sum of all the small arc lengths, $$dS$$, along the circle $$C$$ gives the circumference of a circle of radius $$1$$. Thus,

$$\int_CdS=2πr=2π$$

and Equation (1) becomes

$$2\int_C=2(2π)=4π.\tag{2}$$

Thus, the area of a tube with a height of $$2$$ and a radius of $$1$$ is $$4π$$ square units. More generally, if we considered any arbitrary tube of height $$L$$ and radius $$r$$, using Equations (1) and (2) the surface area of that rube would be given by

$$L\int_CdS=L(2πr).$$

Problem 2: Suppose that a proton $$P_2$$ is moved against an electric field generated by a stationary proton $$P_1$$. If the initial separation distance between $$P_1$$ and $$P_2$$ is $$r_0$$ and the final separation distance is $$r$$, find the work done on $$P_2$$ by the electric force as $$P_2$$ moves from $$r_0$$ to $$r$$.

Figure 2: A proton $$P_2$$ seperated by an initil seperation distance $$r_0$$ from a station proton $$P_1$$ moves along an arbitrary path until it reaches a seperation distance of $$r$$. According to Column's law, the proton $$P_1$$ will exert an electric force and, hence, also do work on the proton $$P_2$$ as it moves from $$r_0$$ to $$r$$. The electric force forms a vector field and by calculuating the line integral along $$P_2$$'s path, we can calculate the amount of work done on $$P_2$$.

There are geometrical applications of line integrals as we learned in the previous problem (such as finding the areas of various different kinds of surfaces); but let's look at some physical applications of line integrals. In this problem, we'll learn bout the application of line integrals in a concept from physics known as work. Suppose that some force $$\vec{F}$$ acts on an object as it moves along some path $$C$$. If $$θ$$ is the angle between the force $$\vec{F}$$ and the object's instantaneous displacement $$d\vec{S}$$ at a point, then the work $$W$$ done on that object is defined as

$$W≡\int_C(Fcosθ)dS=\int_CF_sdS,\tag{1}$$

where $$F_s=Fcosθ$$ is the component of the force $$\vec{F}$$ acting in the same direction as the object's instantaneous displacement at a point. Let's say that a proton $$P_2$$ an initial distance $$r_0$$ away from the stationary proton $$P_1$$ moves along any path $$C$$ until it reaches its final position at a distance $$r$$ away from the stationary proton $$P_1$$. The force $$F_e$$ acting on $$P_2$$ is the electric force which is given by Column's law:

$$F_e=k_e\frac{Q_P^2}{r^2},\tag{2}$$

where $$k_e$$ is a constant, $$Q_P$$ is the electrical charge of a proton, and $$r$$ is the separation distance between two charged particles. Substituting Equation (2) into (1), we find that the work $$W$$ done on $$P_2$$ by the electric force is

$$W=\int_Ck_e\frac{Q_P^2}{r^2}dS.\tag{3}$$

Since the arc length $$dS$$ is equal to an infinitesimal change in $$P_2$$'s radial distance ($$dr$$) away from $$P_1$$, we can rewrite Equation (3) as

$$\int_{r_0}^rk_e\frac{Q_P^2}{r^2}dr.\tag{4}$$

As we discussed in the introductory lesson on line integrals, all line integrals are calculated as a definite integral with respect to a single variable. We can see that Expression (4) is just a plain old definite integral which is straight forward to calculate. Using Expression (4), we can calculate that the work $$W$$ done on $$P_2$$ by the electric force $$F_e$$ is

$$W=\int_{r_0}^rk_e\frac{Q_P^2}{r^2}dr=k_eQ_P^2\int_{r_0}^r\frac{1}{r^2}dr=k_eQ_P^2\biggl[-\frac{1}{r}\biggr]_{r_0}^r=k_eQ_P^2\biggl[\frac{1}{r}\biggr]_r^{r_0}=k_eQ_p^2\biggl(\frac{1}{r_0}-\frac{1}{r}\biggr).$$

Thus, the work done on a proton as it moves towards another stationary proton is

$$W=k_eQ_p^2\biggl(\frac{1}{r_0}-\frac{1}{r}\biggr).\tag{5}$$

Notice that Equation (5) depends on only the initial and final positions of the proton. Thus, the same amount of work would have been done on the proton if it traveled along any arbitrary path from $$r_0$$ to $$r$$. Since the amount of work done by the electric force $$F_e$$ does not depend on a charged particle's path through space, we call the electric force a path-independent or conservative force.

Problem 3: Suppose that a mass $$m$$ is attached to a string of length $$L$$ and is at an intial vertical height of $$h$$. As the mass swings to its final position, its path through space is the curve $$C$$ and its angular displacement is $$θ$$. Find the work $$W$$ done by the force of gravity.

Figure 3: A mass $$m$$ is attached to the end of a rope of length $$L$$. The mass falls and travels a distance $$s$$ with a total angular displacement of $$θ$$.

The gravitational force acting on $$m$$ is $$F_g=-mg$$ and always acts in the "down" direction. As you can see from the free-body diagram of $$m$$ in Figure 3, the component of the gravitational force acting in the direction of $$m$$'s displacement is $$F_{g,s}=-mgsinθ$$. Substituting this result into the definition of work, we find that

$$W=\int_CF_{g,s}dS=-mg\int_CsinθdS.\tag{6}$$

The arc length $$S$$ along a section of a circle is equal to $$rθ$$ where $$r$$ is the radius of the circle and $$θ$$ is the angle subtending $$S$$. Thus, the arc length $$dS$$ is given by

$$dS=Ldθ.$$

Substituting $$dS=Ldθ$$ into Equation (6) and solving the integral, we have

$$W=-mgL\int_{θ_i}^{θ_f}sinθdθ=-mgL\biggl[cosθ\biggr]_{θ_i}^{θ_f}=-mgL\biggl[cosθ\biggr]_{θ_f}^{θ_i}=-mgL(cosθ_i-cosθ_f).$$

If we choose $$θ_i$$ as our reference angle (this is equivalent to letting $$θ_i=0$$), then $$cosθ_i=cos0=1$$ and the equation above simplifies to

$$W=-mgL(1-cosθ).\tag{7}$$

As you can see geometrically from Figure 3, $$L-Lcosθ=h$$. Substituting this result into Equation (7), we have

$$W=-mgh.\tag{8}$$

Since Equation (8) only depends on the initial and final positions of $$m$$ and not on its path $$C$$, the gravitational force is conservative force.