Ask Questions, Get Answers

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

Find the perimeter of the circle with radius $a$.

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 $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
Equation of the circle is $ x^2+y^2=x^2$ .
By symmetry, the arc length (circumference ) is 4x length of are in the arc length in the first quadrent
$L=4 \int \limits _0^a \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2 \normalsize dx}$
Step 2:
Now $x^2+y^2=a^2$
$\therefore \;2x+2ydy=0$
Step 3:
$1+\bigg(\large\frac{dy}{dx}\bigg)$$=1+ \large\frac{x^2}{y^2}$
Step 4:
$L= 4 \int \limits_0^a \large\frac{a}{\sqrt {a^2-x^2}}\;$$dx$
$\quad= 4a \sin^{-1} \large\frac{x}{a}\bigg]_0^a $
$\quad=4a \times \large\frac{\pi}{2}$
$\quad=2 \pi a \;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