Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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.

Can you answer this question?

1 Answer

0 votes
  • 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
$s= \int \limits_0^a 2 \pi y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}dx$
$2y. \large\frac{dy}{dx}$$= 4a$
$1+\bigg(\large \frac{dy}{dx}\bigg)^2$$ =1+ \large\frac{4a^2}{y^2}$
$\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] $
$\qquad= \large\frac{8 \pi}{3}$$ a^2[2 \sqrt 2-1]\;sq.units $
answered Aug 17, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App