Problem 61 - Essential Calculus

Question

In Exercises 61 and \(62,\) use the Second Theorem of Pappus, which is stated as follows. If a segment of a plane curve \(C\) is revolved about an axis that does not intersect the curve (except possibly at its endpoints), the area \(S\) of the resulting surface of revolution is given by the product of the length of \(C\) times the distance \(d\) traveled by the centroid of \(C\). A sphere is formed by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}\) about the \(x\) -axis. Use the formula for surface area, \(S=4 \pi r^{2},\) to find the centroid of the semicircle \(y=\sqrt{r^{2}-x^{2}}\)

Step-by-Step Solution

Answer

The centroid of the semicircle \(y=\sqrt{r^{2}-x^{2}}\) is at \( \bar{y} = \frac{2r}{\pi} \)

Step 1: Formulate the Problem

The problem wants the centroid or geometric center of the semicircle \(y = \sqrt{r^{2}-x^{2}}\). This is a semicircle with radius \(r\) centered at the origin. The centroid of any symmetrical 2D shape like this semicircle lies along the axis of symmetry, which is the y-axis in this case. So the aim is to find the y-coordinate of the centroid

Step 2: The centroid formula

The centroid of a curve/line segment is given by the formula: \( \bar{y} = \frac{1}{A} \int y ds \) where \( A \) is the area under the curve and \( ds \) is a tiny segment along the curve

Step 3: Calculate the Area

The area of the semicircle, \(A\), is \( \frac{1}{2} \pi r^{2} \)

Step 4: Set up the integral

Since you're integrating along the curve, you need to express \( ds \) in terms of \( dx \). To do that, use the Pythagorean theorem factoring in that \( y = \sqrt{r^{2}-x^{2}} \), resulting in \( ds = \sqrt{1 + (\frac{-x}{\sqrt{r^{2}-x^{2}}})^{2}} dx \), which simplifies down to \( ds = \sqrt{1 + (\frac{x^{2}}{r^{2}-x^{2}})} dx \), and simplifies further down to \( ds = dx \) because \( x^{2} + (r^{2} - x^{2}) = r^{2} \)

Step 5: Evaluate the integral

With \( ds \) simplified down to \( dx \), the formula for \( \bar{y} \) now reads \( \bar{y} = \frac{1}{A} \int y dx \) where \( A = \frac{1}{2} \pi r^{2} \) is the area from Step 3 and \( y = \sqrt{r^{2}-x^{2}} \) is the semicircle formula. Depending on the quadrant, x ranges from -r to r. Therefore, \( \bar{y} = \frac{1}{\frac{1}{2} \pi r^{2}} \int_{-r}^{r} \sqrt{r^{2}-x^{2}} dx \). This evaluates to \( \bar{y} = \frac{2}{\pi r} \) times the area of semicircle leading to \( \bar{y} = \frac{2}{\pi r} \times \frac{1}{2}\pi r^{2} = \frac{4r}{4\pi} = \frac{2r}{\pi} \)

Step 6: Final answer

The final answer is the y-coordinate of the centroid, \( \bar{y} = \frac{2r}{\pi} \)

Key Concepts

Centroid of geometric shapes

The centroid of a geometric shape is its center of mass, which serves as a point where it could be perfectly balanced. Think of it like the average position of all the points in a shape. For symmetrical shapes like semicircles, the location of the centroid is often along the axis of symmetry. In our exercise, the semicircle is centered at the origin along the x-axis, meaning its highest point lies on this axis as well.

To find the exact y-coordinate of the centroid, we use the centroid formula for curves. Specifically, for a semicircle described by the equation \( y = \sqrt{r^2-x^2} \), the x-coordinate of the centroid is zero due to the shape's symmetry, while the y-coordinate is calculated using an integral because the shape extends into space vertically.

This y-coordinate of the centroid can be found using the formula \( \bar{y} = \frac{1}{A} \int y \, ds \), where \( A \) is the area of the semicircle, and \( ds \) represents an infinitesimally small segment of the curve.

Surface area of revolution

When a curve is revolved around an axis, it creates a 3D surface. Imagine spinning a shape around a pivot point to form objects like spheres or toruses. The surface area of this new shape can be significant to understand, especially in physics and engineering.

In our scenario, we use the Second Theorem of Pappus to find the surface area of a semicircle revolving around the x-axis to form a sphere. The theorem states that the surface area is the product of the curve's length and the distance traveled by its centroid around an axis. This creates a perfect sphere when a full circle revolves, revealing the inherent symmetry and simplicity of geometric transformation through rotation.

This theorem applies to the problem since the expression for the surface area of a full sphere is already known as \( S = 4 \pi r^2 \). Understanding this rotation and transformation of shapes gives insight into both geometry and calculus principles.

Integration techniques

Integration is a crucial technique in calculus to find areas, volumes, and other quantities that accumulate over a continuum. It transforms the summation of infinitesimally small quantities into a finite result. In the original exercise, integration is used to locate the centroid of the semicircle.

To integrate along the curve, we need to express the differential arc length, \( ds \), in terms of the basic differential \( dx \). This incorporation of the curve's equation, \( y = \sqrt{r^2-x^2} \), and then simplifying it to \( ds = dx \) makes the integration tractable. This simplification relies on symmetry and properties of the semicircle, which keeps calculations both straightforward and insightful.

The integration process then combines with the area of the shape to yield the centroid by folding the curve over its integral limits , as well as factoring in symmetry and radial constraints.

Geometry of semicircles

Semicircles are half of a circle and exhibit several fascinating geometric properties. They are perfectly symmetrical along their diameter, meaning the line passing through the center from one side to the other divides the semicircle into two equal parts.

In the exercise, the semicircle is represented by \( y = \sqrt{r^2-x^2} \), forming part of a circle derived from the Pythagorean theorem. The geometry of semicircles plays an essential role when we examine rotation to form spheres or analyze balance for calculating centroids. Their symmetry simplifies many mathematical calculations, as seen in the problem where a semicircle, once revolved around the x-axis, forms a perfect sphere.

Knowing how semicircles work helps to make sense of rotations into 3D shapes or find useful properties like centroids. All these geometrical insights stem from understanding these important basic shapes and their mathematics.