Question

Find the surface area of the band of the sphere generated by revolving the arc of the cicle
x2+y2=r2
lying above the interval [-a,a],a,r

Answer 1

Answer on Question #45961 – Math – Integral Calculus

Find the surface area of the band of the sphere generated by revolving the arc of the circle $x^2 + y^2 = r^2$ lying above the interval [-a,a],a,r

Solution.

In the case when $f(x)$ is positive and has a continuous derivative, the surface area of the surface generated by revolving the curve $y = f(x)$, $x_1 \leq x \leq x_2$ about the $x$-axis is

$$S = 2 \pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx.$$

In our case: $y = \sqrt{r^2 - x^2}$, $x_1 = -a$, $x_2 = a$,

$$\frac{dy}{dx} = \frac{1}{2} (r^2 - x^2)^{-\frac{1}{2}} (-2x) = -\frac{x}{\sqrt{r^2 - x^2}}.$$

Thus,

$$\begin{aligned} S &= 2 \pi \int_{-a}^{a} \sqrt{r^2 - x^2} \sqrt{1 + \frac{x^2}{r^2 - x^2}} \, dx = 2 \pi \int_{-a}^{a} \sqrt{r^2 - x^2} \frac{r}{\sqrt{r^2 - x^2}} \, dx = \\ &= 2 \pi r \int_{-a}^{a} dx = 2 \pi r x \big|_{x = -a}^{x = a} = 2 \pi a r + 2 \pi a r = 4 \pi a r. \end{aligned}$$

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