Question

A sphere of radius \(r\) is generated by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}\) about the \(x\) -axis. Verify that the surface area of the sphere is \(4 \pi r^{2}\)

Short answer

The surface area of a sphere generated by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}\) about the \(x\)-axis is \(4 \pi r^{2}\)

Step-by-step solution

Understanding the problem

A sphere of radius \(r\) is generated by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}\) about the \(x\)-axis. So first off, one should understand this sphere has symmetry about the \(x\)-axis, with \(r\) defined as the square root of the difference between the square of the radius and the square of the \(x\)-axis.

Setting up the Surface Area Integral

The formula for calculating surface area of revolution (when rotating around the \(x\)-axis) is \(A = 2\pi \int_{a}^{b}y \sqrt{1 + ((dy/dx)^{2})}dx\), where \(y\) represents the function being rotated, \(dy/dx\) is the derivative of that function and \(a\) and \(b\) are the bounds of \(x\). In this case, \(y = \sqrt{r^{2} - x^{2}}\), \(-r \leq x \leq r\) and \(dy/dx = -x/ \sqrt{r^{2} - x^{2}}\). Hence the Integral for Surface Area becomes \(A = 2\pi \int_{-r}^{r}\sqrt{r^{2} - x^{2}} \sqrt{1 + ((-x/ \sqrt{r^{2} - x^{2}})^{2})}dx\)

Simplifying the Integral

Solving the square root inside the integral yields the integrand as \(\sqrt{r^{2}-x^{2}} \sqrt{1+ x^{2}/(r^{2}-x^{2})}\). Further simplification of the \(1+ x^{2}/(r^{2}-x^{2})\) results in \(r\), so the simplified integral is \(A = 2\pi \int_{-r}^{r}r dx\)

Evaluating the Integral

The anti-derivative of the integrand with respect to \(x\) is \(rx\). So the evaluation of the integral from \(-r\) to \(r\) is \(2\pi [r*r -(-r)*r] = 4 \pi r^{2}\).

Final answer

So, the surface area of the sphere obtained from revolving \(y=\sqrt{r^{2}-x^{2}}\) around the \(x\)-axis is indeed \(4 \pi r^{2}\).