MATH 146 – Calculus II
December 23, 2025
Two magnitudes balance at distances reciprocally proportional to their magnitudes.
In other words, \(\frac{d_1}{d_2} = \frac{m_2}{m_1} \Longleftrightarrow d_1 m_1 = d_2 m_2\).
Archimedes showed that:
Given any region bounded by \(f(x)\) and \(g(x)\), can we find the coordinates \((\overline{x}, \overline{y})\) of its center of gravity?
If masses \(m_1, m_2, \dots, m_n\) are suspended on a balance beam from points \(x_1, x_2, \dots, x_n\), then the system will be in balance when suspended from the point \(\overline{x}\) such that:
\[ \sum_{i=1}^n m_i (\overline{x} - x_i) = 0, \]
where:
\[ \begin{aligned} \overline{x} = \dfrac{\displaystyle\sum_{i=1}^n m_i x_i}{\displaystyle\sum_{i=1}^n m_i}. \end{aligned} \]
\[ \begin{aligned} \overline{x} = \frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i} \quad\Longrightarrow \quad \overline{x}_P = \frac{\sum_{i=1}^n x_i^* (f(x_i^*) - g(x_i^*)) \Delta x_i}{\sum_{i=1}^n (f(x_i^*) - g(x_i^*)) \Delta x_i} \end{aligned} \]
If we send the width of the intervals to 0, we can rewrite these as intervals.
\[ \begin{aligned} \overline{x}_P &= \lim\limits_{\Delta x \to 0} \frac{\sum_{i=1}^n x_i^* (f(x_i^*) - g(x_i^*)) \Delta x_i}{\sum_{i=1}^n (f(x_i^*) - g(x_i^*)) \Delta x_i} \\[1em] &= \frac{\displaystyle\int_a^b x\left(f(x) - g(x)\right) \, dx}{\displaystyle\int_a^b f(x) - g(x) \, dx} \end{aligned} \]
\[ \begin{aligned} \overline{y} = \frac{\sum_{i=1}^n m_i y_i}{\sum_{i=1}^n m_i} \,\Longrightarrow \, \overline{y}_P = \frac{\sum_{i=1}^n \left( \frac{f(x_i^*) + g(x_i^*)}{2} \right) \left( f(x_i^*) - g(x_i^*) \right) \Delta x_i}{\sum_{i=1}^n (f(x_i^*) - g(x_i^*)) \Delta x_i} \end{aligned} \]
If we send the width of the intervals to 0, we can rewrite these as intervals.
\[ \begin{aligned} \overline{y}_P &= \lim\limits_{\Delta x \to 0} \frac{\sum_{i=1}^n \left( \frac{f(x_i^*) + g(x_i^*)}{2} \right) \left( f(x_i^*) - g(x_i^*) \right) \Delta x_i}{\sum_{i=1}^n (f(x_i^*) - g(x_i^*)) \Delta x_i} \\[1em] &= \frac{1}{2} \left( \frac{\displaystyle\int_a^b f(x)^2 - g(x)^2 \, dx}{\displaystyle\int_a^b f(x) - g(x) \, dx} \right) \end{aligned} \]
If a region \(R\) is bounded by the graphs of \(f(x) > g(x)\) on the interval \([a, b]\), then the center of gravity of \(R\) can be found at the point with coordinates \((\overline{x}, \overline{y})\), where:
\[ \begin{aligned} \overline{x} &= \frac{\displaystyle\int_a^b x\left(f(x) - g(x)\right) \, dx}{\displaystyle\int_a^b f(x) - g(x) \, dx} \\[0.5em] \overline{y} &= \frac{\displaystyle\int_a^b f(x)^2 - g(x)^2 \, dx}{2 \displaystyle\int_a^b f(x) - g(x) \, dx} \end{aligned} \]
Example: Find the center of gravity of the region bounded by the curves \(y=1\) and \(y=x^2\).
Example: Find the center of gravity of the region bounded by the curves \(y=\sqrt{x}\) and \(y=x^3\).
Given a curve \(f(x) \ge 0\) over the interval \([a, b]\), a volume of revolution is the 3D figure obtained by revolving the region between \(f(x)\) and the \(x\)-axis about an axis.
Example: \(f(x) = \sin(x)\) on the interval \(x \in [0, \pi]\).
How do we calculate the volume of these?
Let \(R\) be a plane region that lies entirely on one side of a line \(l\) in the plane. If \(R\) is rotated about \(l\), then the volume of the resulting solid is the product of the area \(A\) of \(R\) and the distance traveled by the centroid of \(R\).
Suppose \(f(x) \ge g(x) \ge 0\) over an interval \([a, b]\). Then the volume of revolution obtained by rotating the region contained by \(f(x)\) and \(g(x)\) over the interval \([a,b]\) around the \(x\)–axis is:
\[ V = \pi \int_a^b f(x)^2 - g(x)^2 \, dx \]
\[ \begin{aligned} V &= (2 \pi \overline{y}) \times A_R \\ &= 2\pi \left( \frac{\displaystyle\int_a^b f(x)^2 - g(x)^2 \, dx}{2 \displaystyle\int_a^b f(x) - g(x) \, dx} \right) \left( \int_a^b f(x) - g(x) \, dx \right) \end{aligned} \]
Example: Find the volume of revolution obtained by revolving the region beneath the curve \(y = x^2\) around the \(x\)–axis with \(0 \le x \le 1\).
Example: Find the volume of revolution obtained by revolving the region beneath the curves \(y = x^2\) and \(y = 1\) around the \(x\)–axis with \(0 \le x \le 1\).
Suppose \(f(x) \ge g(x) \ge 0\) over an interval \([a, b]\). Then the volume of revolution obtained by rotating the region contained by \(f(x)\) and \(g(x)\) over the interval \([a,b]\) around the \(y\)–axis is:
\[ V = 2\pi \int_a^b x \left( f(x) - g(x) \right) \, dx \]
\[ \begin{aligned} V &= (2 \pi \overline{x}) \times A_R \\ &= 2\pi \left( \frac{\displaystyle\int_a^b x\left(f(x) - g(x)\right) \, dx}{\displaystyle\int_a^b f(x) - g(x) \, dx} \right) \left( \int_a^b f(x) - g(x) \, dx \right) \end{aligned} \]
Example: Find the volume of revolution obtained by revolving the region beneath the curve \(y = x^2\) around the \(x\)–axis with \(0 \le y \le 1\).
Example: Find the volume of revolution obtained by revolving the region beneath the curves \(y = x^2\) and \(y = 1\) around the \(y\)–axis with \(0 \le x \le 1\).