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MTH 252Z Integral Calculus Fall 2025

Richard Zhao

Section 6.2 Setting up Integrals: Volume, Density, Average Value

In this section, we will discuss how to find the volume of a solid and the average value of a function using integrals.
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Subsection Volume by Cross Section

Let’s give a deeper consideration of what the term volume means geometrically. Imagine we can slice through the solid vertically into millions and millions of pieces with infinitesimally small thickness. Then we can find the area of each cross section, multiply it with the infinitesimally small thickness, and add up all the pieces.
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Figure 6.2.1. Vertical Cross Section of a Solid
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If we assume the vertical cross section to have an area of \(A(x)\) and each slice have the infinitesimally small thickness of \(dx\text{,}\) then the volume of the solid is the sum of the volume of all the slices. That is,
\begin{equation*} V = \int_a^b A(x)\, dx \end{equation*}
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Similarly, if we are slicing the solid horizontally, then the infinitesimally small thickness will be \(dy\text{.}\)
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Figure 6.2.2. Horizontal Cross Section of a Solid
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If we assume, similarly, that the horizontal cross section to have an area of \(A(y)\) and each slice have the infinitesimally small thickness of \(dy\text{,}\) then the volume of the solid is the sum of the volume of all the slices. That is,
\begin{equation*} V = \int_a^b A(y)\, dy \end{equation*}
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Now we can make a fancy theorem out of it.
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Observe that there is a difference between which variable, \(x\) or \(y\text{,}\) we are using in the formula depending on the orientation of the slices. One way to remember this is by determining the differential first. If the slices are vertical, then the thickness is some horizontal distance (observe the diagram in Figure 6.2.1), which means the thickness is \(dx\text{.}\) Then the variable in the integral should be \(x\text{.}\) This is why we want to find the area of the cross sections with respect to \(x\text{.}\)
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You can make a similar argument on why using the variable \(y\) when the slices are horizontal (by observing the diagram in Figure 6.2.2).
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Example 6.2.4.

Find the volume of the solid whose base is the semi-circle \(y = \sqrt{9 - x^2}\) and the cross section perpendicular to the \(x\)-axis are squares.
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Hint.
You should imagine what the solid looks like based on this description. Richard understands that this may be super difficult (he doesn’t have a good 3-dimensional visualization skill so he understands how difficult this can be, but he did have students in the past who are super good at this...so everything is possible!).
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To help you visualize this solid, Richard included a Desmos applet below:
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Now that you have an idea of what the solid looks like, how do we determine the area of the cross section and the interval of which the \(x\)-values is in?
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Solution.
Since the slices are vertical, we can find the volume of the solid using the formula
\begin{equation*} \text{Volume} = \int_a^b A(x)\, dx \end{equation*}
So the two things we want to find are (1) the area of the cross section, and (2) the interval in which the \(x\)-values are in.
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Let’s figure out the area of the cross section first. By the problem, we know that the cross sections are squares, whose area can be computed by squaring the sides. So the question now becomes what the side of the squares are?
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By the Desmos applet in the hint, the base of the square is the height of the function \(y = \sqrt{9 - x^2}\text{,}\) which is the function value. Then the area of the cross section is
\begin{equation*} A(x) = \left(\sqrt{9 - x^2}\right)^2 = 9 - x^2 \end{equation*}
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Now we will determine the interval in which the \(x\)-values live in. Graphically, we can observe that the \(x\)-values will be within the interval of \([-3,3]\text{.}\)
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Putting these two pieces together, the volume of the solid is
\begin{align*} \text{Volume} \amp= \int_{-3}^3 \left(9 - x^2\right)\, dx \\ \amp= \left(9x - \frac{x^3}{3}\right)\bigg|_{-3}^3 \\ \amp= \left(9(3) - \frac{(3)^3}{3}\right) - \left(9(-3) - \frac{(-3)^3}{3}\right) \\ \amp= 18 - (-18) \\ \amp= 36 \qquad \text{unit}^3 \end{align*}
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Subsection Average Value

In statistics, we can find the average by adding up all the data values and dividing the sum by the number of data values. Symbolically, the average in statistics is
\begin{equation*} \frac{a_1 + a_2 + \cdots + a_N}{N} = \frac{1}{N}\sum_{i = 1}^N a_i \end{equation*}
But this formula is only useful if we have finite number of the data values. What if we want to find the average value of a continuous function within some interval? We will have infinite function values to deal with...
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One way to work around this issue is to borrow the idea of the right-endpoint approximation back in Section 5.1.
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Let’s say we want to find the average value of a continuous function \(f\) over some closed interval \([a,b]\text{.}\) We will first find the right-endpoint approximation as follows:
\begin{align*} R_N \amp= \frac{b - a}{N} \left(f(x_1) + f(x_2) + \cdots + f(x_N)\right)\\ \amp= (b - a)\cdot \frac{f(x_1) + f(x_2) + \cdots + f(x_N)}{N} \end{align*}
Observe that the fraction on the right-hand side of the equation looks a lot like the average value in the statistical sense! So let’s isolate it!
\begin{equation*} \frac{f(x_1) + f(x_2) + \cdots + f(x_N)}{N} = \frac{1}{b - a}\cdot R_N \end{equation*}
If we want to sum up all the function values, then \(N\to \infty\) here. Imagine we are taking the limit on both sides of the equation as \(N\to \infty\text{,}\) we have
\begin{align*} \lim_{N\to \infty} \left(\frac{f(x_1) + f(x_2) + \cdots + f(x_N)}{N}\right) \amp= \lim_{N\to \infty} \left(\frac{1}{b-a}\cdot R_N\right) \\ \amp = \frac{1}{b-a}\cdot \lim_{N\to \infty} R_N \end{align*}
Hmm but what is the right-endpoint approximation when \(N\to \infty\text{...}\) That is the definite integral! (Recall we defined the definite integrals as the limit of the Riemann sum, and \(R_N\) is just one special Riemann sum!). That is, the average value of a continuous function can be computed by
\begin{equation*} \text{Average Value} = \frac{1}{b-a}\int_a^b f(x)\, dx \end{equation*}
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Example 6.2.5.

The temperature (in degrees Celsius) at time \(t\) (in hours) in an art museum varies according to
\begin{equation*} T(t) = 20 + 5\cos\left(\frac{\pi}{12}t\right) \end{equation*}
Find the average temperature in the museum over a day.
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Hint.
Observe the temperature function is a continuous function, so it is impossible to find all the temperature, adding them up, and divide the sum by the number of temperature values.
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Instead, we just derived a fancy formula to find the average value of a continuous function over an interval. Let’s use this formula!
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What is the function and what is the interval of interest?
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Solution.
The function we are looking at here is the temperature function
\begin{equation*} T(t) = 20 + 5\cos\left(\frac{\pi}{12}t\right) \end{equation*}
We care about the average temperature over a day, which has 24 hours, so the interval of interest is \([0,24]\text{.}\)
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Now we can plug things into the average value formula!
\begin{align*} \text{Average Value} \amp= \frac{1}{24}\int_0^{24} \left(20 + 5\cos\left(\frac{\pi}{12}t\right)\right)\, dt \\ \amp= \frac{1}{24}\left(20t + 5\cdot \frac{12}{\pi}\sin\left(\frac{\pi}{12} t\right)\right)\bigg|_0^{24} \\ \amp= \frac{1}{24}\left(20\cdot 24 + \frac{60}{\pi}\sin\left(\frac{\pi}{12}\cdot 24\right)\right) \\ \amp= 20 \end{align*}
Hence, the average temperature of the museum over a day is \(20 \text{C}^\circ\text{.}\)
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Observe there are formulas you can apply directly in this section. To make sure you are familiar with the formulas, spend some time working through some practice problems, especially the volume idea by summing up the volume of smaller slices. The two big sections in this chapter, section 6.3 and 6.4, will rely on this idea of the volume.
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