Browse Questions

# Find the surface area of the solid generated by revolving the arc of the parabola$\;y^{2}=4ax$, bounded by its latus rectum about $x$- axis.

• 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 $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
To find the surface area of the solid obtained by rotating the shaded area (between $x=0,x=a$) of the parabola about the x axis
$s= \int \limits_0^a 2 \pi y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}dx$
$y^2=4ax$
$2y. \large\frac{dy}{dx}$$= 4a \large\frac{dy}{dx}=\large\frac{2a}{y} 1+\bigg(\large \frac{dy}{dx}\bigg)^2$$ =1+ \large\frac{4a^2}{y^2}$
$\qquad=1+\large\frac{4a^2}{4ax}=\frac{x+a}{x}$
$\therefore s= 2 \pi \int \limits_0^a 2 \sqrt a \sqrt x \sqrt {\large\frac{x+a}{x}} $$dx \qquad= 4 \pi \sqrt a \int \limits_o^a (x+a)^{\large\frac{1}{2}}$$ dx$
$\qquad= 4 \pi a^{\large\frac{1}{2}}.\large\frac{2}{3}$$(x+a)^{\large\frac{3}{2}} \bigg]_0^a \qquad= \large\frac{8 \pi}{3} a^{\large\frac{1}{2}}$$\bigg[(2a)^{\large\frac{3}{2}}-a^{\large\frac{3}{2}} \bigg]$