Question

A cone is generated when rhe region is bounded by the line h=y and the vertical line x=0 and n=r is rotated about the x-axis. use the pappus theorem to show that the surface area(s) of the cone is giveb by S=πr√r^2 + h^2

Answer 1

If a plane curve is rotated about an axis in its plane, but which does not cross the curve, the area swept out equals the length times the distance moved by the centroid.

$$x=0, y=h$$

The equation of the line that forms the hypotenuse is

$$y=-\dfrac{h}{r}x+h$$

Consider a line of length $L.$ Its centroid is at a distance $\bar{y}$  from the x axis. Rotate the line through $360\degree$about the x axis. The distance moved by the centroid is

$$2\pi\cdot\bar{y}=2\pi\cdot(\dfrac{1}{2}h)=\pi h$$

By the Pythagorean Theorem $L=\sqrt{r^2+h^2}.$ The surface area of the cone swept out is

$$S=L\times(2\pi\cdot\bar{y})=\pi h\sqrt{r^2+h^2}$$