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 \(y=\sqrt{r^{2}-x^{2}}\)

Short answer

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

Step-by-step solution

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

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

Calculate the Area

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

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 \( x^{2} + (r^{2} - x^{2}) = r^{2} \)

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 \( \bar{y} = \frac{2}{\pi r} \times \frac{1}{2}\pi r^{2} = \frac{4r}{4\pi} = \frac{2r}{\pi} \)

Final answer

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