Browse Questions

# Prove that the curved surface area of a sphere of radius $r$ intercepted between two parallel planes at a distance $a$ and $b$ from the centre of the sphere is $\;2\pi r(b-a)$ and hence deduct the surface area of the sphere,$(b>a)$

Toolbox:
• Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx$ or $\int \limits _a^b y dx$
• If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
A sphere is obtained when the region between $x=-r,x=r$ of the circle $x^2+y^2=r^2$ in the first two quadrents is rotated about the $x-axis$
Consider the section by the sphere obtained by rotating the hashed area between $x=a,x=b( b >a)$
The surface area $S=2 \pi \int \limits_a^b y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}dx$
Now $x^2 +y^2 =r^2$
$\therefore 2x+2y \large\frac{dy}{dx}$$=0=>\large\frac{dy}{dx}=\frac{-x}{y} 1+\bigg(\large\frac{dx}{dy}\bigg)^2$$=1+\large\frac{x^2}{y^2}$
$\qquad= \large\frac{x^2+y^2}{y^2}=\frac{r^2}{r^2-x^2}$
$y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}=\sqrt {r^2-x^2}. \large\frac{r}{\sqrt {r^2-x^2}}$$=r$
$\therefore S=2 \pi \int \limits_a^b r dx=2 \pi r (x)_a^b$
$\qquad= 2 \pi r (b-a) \;sq.units$
The surface of the sphere is obtained by substituting $a=-r,b=r$
$\therefore S= 2 \pi r (2r) =4 \pi r^2 sq.units$