# 5.5: Applications of Integration

**At Grade**Created by: CK-12

## Area Between Two Curves

When students first started taking definite integrals the interesting case of what was actually being represented was illustrated by\begin{align*}\int\limits_{-3}^{3}x^3 dx=0\end{align*}

It's worth looking at a simple case for the first question. Find the area between \begin{align*}f(x)=x^3-2x^2+6\end{align*}

One curve is completely above the axis, the other is below. Ask the students, What do you think will happen? To calculate:

\begin{align*} \int\limits_{-1}^{2} (x^3 - 2x^2 + 6) - (x^ - 4x - 8)dx & = \int\limits_{-1}^{2} x^3 - 3x^2 + 4x + 14 dx \\
& = \frac{1}{4} x^4 - x^3 + 2x^2 + 14x \big|^2_{-1} \\
& = \left (\frac{1}{4} (2)^4 - (2)^3 + 14(2) \right ) - \left (\frac{1}{4} - (-1)^4 - (-1)^3 + 2 (-1)^2 + 14(-1) \right ) \\
& = 36 + 10.75 \\
& = 46.75\end{align*}

It would be useful to have some groups working on the problem this way, and other working on the area under the top curve to the axis, taking the negative area of the curve below the \begin{align*}x\end{align*}

The problem is that the curve “on top” changes. The general form isn't really “the area between two curves” in the same way that the definite integral is not “the area between the curve and the axis.” The equations will need to be solved to find the point of intersection, and then two different definite integrals will need to be taken. A computer solver give that the intersection happens at \begin{align*}x=-1.39\end{align*}

\begin{align*}\int\limits_{-2}^{-1.39}(x^3 - 2x^2 + 6) - (x^2 - 4x - s-8) dx + \int\limits_{-2}^{-1.39} (x^3 - 2x^2 + 6) - (x^2 - 4x - s-8) dx\end{align*}

## Volumes

It is valuable have a conceptual understanding of the idea that cross sectional areas added together allows for the calculation of volumes. There are more methods and formulas than one can reasonably remember, although some common, or maybe difficult ones, are worth the time. There are many questions outside of these forms, however, that are favorites on many tests. One that frequently gets chosen is asking for the volume of the solid that has a specified base, with a particular shape above that base. Here is an example:

What is the volume of the solid whose base is bounded by \begin{align*}e^x,x=0,x=1\end{align*}

A picture is very helpful in organizing all the information. The first order of business is to figure out what the area is that is needed to iterate to get the volume requested. The half circles that are shaded darker are the area in question, so they are what we need to figure out the expression for the area of those shapes next.

Since they are semi-circles, the diameter is going to be the length across the bottom from the axis to the curve \begin{align*}e^x\end{align*}

\begin{align*}\int\limits_{0}^{1} \frac{1}{2} \pi \left (\frac{e^x}{2} \right )^2 dx = \frac{\pi}{8} \int\limits_{0}^{1} e^x dx = \frac{\pi}{8} (e-1) \approx .675\end{align*}

It's worth making it into a mantra: “Find volumes by integrating areas for the length of the solid.”

## The Length of a Plane Curve

Often times lines, especially those modeling particle movement in a \begin{align*}2-\end{align*}

Parametric arc length: \begin{align*}A(t) = \int\limits_{\alpha}^{\beta} \sqrt{\left (\frac{dx}{dt} \right )^2 + \left (\frac{dy}{dt} \right )^2} dt\end{align*}

Something students are likely to recognize is that very few functions work very well for the rectangular arc length formula. The combination of taking a derivative, squaring it, adding a term and then taking the integral of the square root of that function rarely results in an easy, if even possible integral to take. Many of these problems will need to have the definite integral approximated by a computer solver, or by using Simpson's rule. I must admit to the reader that I spent many hours searching past notes and texts to find an even remotely interesting parametric function that is a possible integral, and none came up. The easiest integral is finding the circumference, or part of the circumference, of a circle using the parameters \begin{align*}x = \sin(t),y = \cos(t), 0 \le t \le c\end{align*} where c is arbitrary, a \begin{align*}c\end{align*} of \begin{align*}2 \pi\end{align*} gives a complete circle. But this can be computed without the arc length formula, and it's not terribly exciting. Here are some fun ones to try if using a computer based solver, largely because they make cool pictures: Find the length of the figure described by the parametric equations:

\begin{align*}x = \cos (3t), && y = \sin(5t), && 0 \le t \le 2\pi \end{align*}

When using a computer solver, the key is to make sure that the derivatives are taken correctly, and that the input syntax is correct.

\begin{align*}\int\limits_{0}^{2 \pi} \sqrt{(- 3 \sin(3t))^2 + (5 \cos (5t))^2} dt \approx 24.6\end{align*}

Find the length of the line described by the parametric equations

\begin{align*}x = t + \sin(2t), y = \cos(t), 0 \le + \le 2 \pi\end{align*}

\begin{align*}\int\limits_{0}^{2 \pi}\sqrt{(1 + 2 \cos(2y))^2 + (- \sin(t))^2} dt \approx 10.8\end{align*}

## Area of a Surface of Revolution

Newton’s Law of Cooling states that the rate of temperature change is equal to the heat transfer coefficient times the surface area times the difference in temperatures. Stated in variables:

\begin{align*}\frac{dQ}{dt} = hA(T_e-T_o)\end{align*}

Since often times in engineering the temperature of the object and environment is fixed, as well as the material, the surface area is the one thing that can be changed to affect the dissipation of heat. If we are designing a heat sink out of aluminum that needs to dissipate at least \begin{align*}200 \;\mathrm{joules}\end{align*} of heat from a device running at \begin{align*}373K\end{align*} in a \begin{align*}298K\end{align*} environment. The heat transfer coefficient of aluminum in air is \begin{align*}25 \;\mathrm{W/m}^2 K\end{align*}. The shape of the heatsink is the surface made by revolving the function \begin{align*}x^3\end{align*} about the \begin{align*}x\end{align*} axis from the origin. Determine the length along the \begin{align*}x-\end{align*}axis needed to dissipate the required energy.

Here, we need to substitute all the information we have into the Law of Cooling function. This is a little different than normal, as we are not asking to compute the area of the surface, but need to state where the limits of integration are to get the proper area needed to conform to the requirements. Because the integral is going to take up some serious space, we should first solve for the total minimum area.

\begin{align*}-207 = 25A(298-373) \rightarrow A=.1104m^2\end{align*}

Now setting the integral equal to this quantity, but leaving the variable we need to solve for in the upper limit:

\begin{align*}.1104 & = \int\limits_{0}^{L} 2 \pi x^3 \sqrt{1 + 9x^4} dx \rightarrow .1104 = \frac{2\pi}{36} \left (\frac{2}{3}(1 + 9x^4 \right )^{\frac{3}{2}} \big|^L_0\\ .9488 & = (1+9(L)^4) - 1 \rightarrow L =.5698\end{align*}

This tells us we need to have extend the surface to at least \begin{align*}.5689m\end{align*} in length to get the required properties.

## Applications from Physics, Engineering and Statistics

These problems are really illustrative of how calculus was developed and the questions that drove the techniques and theorae learned thus far. Problems that have natural or applied motivation often do not work as “cleanly” as the types of packaged problems typically presented in textbooks for practice. There are a few tools that are helpful in navigating these problems.

- Always keep track of vectored quantities. While it is sometimes a bit of extra work to make sure all the signs are set up in the correct manner, one nice result of doing careful work up front is that the answer falls with the correct sign with only doing the correct math.
- When it doubt, write all the units, all the time. Sometimes the units illustrate the next step and can keep you going when stuck. For example: finding quantities like work involves multiplying two other quantities. If you have force as a function of distance, then the product of the two is area, indicating that an integral is called for.
- Use significant space for work. Some problems or formulae may use odd numbers or expressions that can be confusing if they are crammed into a small space. I am thinking specifically about the standard normal distribution, which has a complex exponent that is easy to get mixed up.
- Draw a picture. Always. The quality of a picture, as well as the labeling of quantities is imperative for keeping track of necessary information, and how the quantities relate.

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