Centroids of Area

The centroid of a plane area (or lamina) is its geometric center. It is the point at which the shape would perfectly balance if it were a thin, flat object of uniform density.

For a simple geometric shape like a rectangle, circle, or triangle, the centroid is often obvious from symmetry. For example:

• The centroid of a rectangle is at the intersection of its diagonals.
• The centroid of a circle is at its center.
• The centroid of a triangle is at the intersection of its medians (one-third of the distance from the midpoint of a side to the opposite vertex).

For more complex shapes, especially those bounded by curves, calculus (specifically integration) is used to determine the coordinates of the centroid, denoted as $(\bar{x}, \bar{y})$.

The centroid is also known as the "center of area" or "center of figure." If the object has uniform density, the centroid coincides with the center of mass.

Consider an area bounded by the curve $y=f(x)$, the x-axis ($y=0$), and the vertical lines $x=a$ and $x=b$. Assuming $f(x) \ge 0$ on $[a,b]$.

1. Area (A): The total area of the region is given by:

$$ A = \int_a^b f(x) \, dx $$

2. Moment about the y-axis ($M_y$): This measures the tendency of the area to rotate about the y-axis. It's calculated by summing the product of elemental areas $dA = f(x)dx$ and their distance $x$ from the y-axis:

$$ M_y = \int_a^b x \cdot f(x) \, dx $$

3. Moment about the x-axis ($M_x$): This measures the tendency of the area to rotate about the x-axis. For an elemental strip of height $f(x)$ and width $dx$, its centroid is at $(x, \frac{f(x)}{2})$. So, the moment of this elemental area $dA = f(x)dx$ about the x-axis is $\frac{f(x)}{2} \cdot f(x)dx$.

$$ M_x = \int_a^b \frac{1}{2} [f(x)]^2 \, dx $$

4. Centroid Coordinates $(\bar{x}, \bar{y})$:

$$ \bar{x} = \frac{M_y}{A} = \frac{\int_a^b x f(x) \, dx}{\int_a^b f(x) \, dx} $$

$$ \bar{y} = \frac{M_x}{A} = \frac{\int_a^b \frac{1}{2} [f(x)]^2 \, dx}{\int_a^b f(x) \, dx} $$

Consider an area bounded by the curve $x=g(y)$, the y-axis ($x=0$), and the horizontal lines $y=c$ and $y=d$. Assuming $g(y) \ge 0$ on $[c,d]$.

1. Area (A): The total area of the region is given by:

$$ A = \int_c^d g(y) \, dy $$

2. Moment about the y-axis ($M_y$): For an elemental strip of width $g(y)$ and height $dy$, its centroid is at $(\frac{g(y)}{2}, y)$. The moment of this elemental area $dA = g(y)dy$ about the y-axis is $\frac{g(y)}{2} \cdot g(y)dy$.

$$ M_y = \int_c^d \frac{1}{2} [g(y)]^2 \, dy $$

3. Moment about the x-axis ($M_x$): This is calculated by summing the product of elemental areas $dA = g(y)dy$ and their distance $y$ from the x-axis:

$$ M_x = \int_c^d y \cdot g(y) \, dy $$

4. Centroid Coordinates $(\bar{x}, \bar{y})$:

$$ \bar{x} = \frac{M_y}{A} = \frac{\int_c^d \frac{1}{2} [g(y)]^2 \, dy}{\int_c^d g(y) \, dy} $$

$$ \bar{y} = \frac{M_x}{A} = \frac{\int_c^d y g(y) \, dy}{\int_c^d g(y) \, dy} $$

The Second Theorem of Pappus (also known as Pappus's Centroid Theorem or Guldinus Theorem) relates the volume of a solid of revolution to the area of the generating plane region and the distance traveled by its centroid.

Statement: The volume $V$ of a solid of revolution generated by revolving a plane area $A$ about an external axis in its plane (that does not intersect the area) is equal to the product of the area $A$ and the distance $d$ traveled by the centroid of the area.

If $\bar{R}$ is the perpendicular distance from the centroid of the area $A$ to the axis of revolution, then the distance traveled by the centroid during one full revolution is $d = 2\pi \bar{R}$.

Thus, the theorem can be stated as:

$$ V = A \cdot d = A \cdot (2\pi \bar{R}) $$

This theorem can be useful for:

• Finding the volume of a solid of revolution if the area $A$ and its centroid's distance $\bar{R}$ from the axis are known.
• Finding the distance $\bar{R}$ of the centroid from an axis if the area $A$ and the volume $V$ of revolution are known: $\bar{R} = \frac{V}{2\pi A}$.

For example, the centroid of a semicircular area of radius $r$ is known to be at a distance $\bar{R} = \frac{4r}{3\pi}$ from its diametral edge. The area of the semicircle is $A = \frac{1}{2}\pi r^2$. If this semicircle is revolved around its diametral edge (the x-axis, if centered at origin), it forms a sphere. Using Pappus's theorem, the volume of the sphere is:

$$ V = \left(\frac{1}{2}\pi r^2\right) \cdot \left(2\pi \cdot \frac{4r}{3\pi}\right) = \left(\frac{1}{2}\pi r^2\right) \cdot \left(\frac{8r}{3}\right) = \frac{4}{3}\pi r^3 $$

Which is the correct formula for the volume of a sphere.