Ask Questions, Get Answers

Home  >>  CBSE XII  >>  Math  >>  Application of Integrals

Find the area of the part of the circle \( x^2+y^2=16\) which is exterior to the parabola \( y^2=6x\).

$\begin{array}{1 1}\large\frac{4}{3}[8\pi-\sqrt 3] sq.units \\ \large\frac{4}{3}[8\pi+\sqrt 3] sq.units \\\large\frac{4}{3}[8\pi-\sqrt 5] sq.units. \\ \large\frac{2}{3}[8\pi-\sqrt 3] sq.units \end{array} $

1 Answer

  • Suppose we are given two curves represented by $y=f(x),y=g(x)$ where $f(x)\geq g(x)$ in [a.b] the points of intersection of the two curves are given by x=a and x=b by taking common values of y from the given equation of the curves.
  • Hence the required area is given by $A=\int _a^b[f(x)-g(x)]dx$
Step 1:
The required area is area of the circle $x^2+y^2=16$ which is extension to the parabola $y^2=6x$
To find the points of intersection,let us solve the given equation.
substituing for $y^2$ in the equation of the circle we get,
$\Rightarrow x^2+6x-16=0$
on factorising we get ,
$\Rightarrow x=-8 \;and\; x=2.$
Step 2:
since for x=-8 we get $y^2=-8$ which is imaginary,we cannot take this value.
Now if x=2 we get y=$\pm 2\sqrt 3$.
Hence the points of intersection are (2,$2\sqrt 3$) and (2,$-2\sqrt 3$).
Now let us take the limits as 0 to 2 for parabola and 2 to 4 for the circle.
The area bounded by the circle and parabola is 2x[area bounded by the parabola and x-axis +area bounded by the circle and x-axis] $A=2\times \begin{bmatrix}\int_0^2\sqrt {6x}dx+\int_2^4\sqrt {16-x^2}dx\end{bmatrix}$
on integrating we get
$A=2\begin{bmatrix}\sqrt 6\frac{x^{\Large\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_0^2+\frac{x}{2}\begin{bmatrix}\sqrt {16-x^2}+\Large\frac{16}{2}\normalsize\sin^{-1}\large\frac{x}{4}\end{bmatrix}_2^4$
Step 3:
on applying limits we get,
$A=\large\frac{4\sqrt 6}{3}$$(2\sqrt 2)+2\begin{bmatrix}0+\frac{16}{2}\sin^{_1}(1)-\sqrt {16-4}+\frac{16}{2}\sin^{-1}\frac{1}{2}\end{bmatrix}$
$A=\large\frac{4\sqrt 6}{3}$$(2\sqrt 2)+2\begin{bmatrix}\frac{8\pi}{2}-4\sqrt 3+8\big(\frac{\pi}{6}\big)\end{bmatrix}$
on simplifying we get,
$A=\large\frac{4}{3}$$[4\sqrt 3+6\pi-3\sqrt 3-2\pi]$
$\;\;\;=\large\frac{4}{3}$$[\sqrt 3+4\pi]$
Area of the circle=$\pi r^2$
$\Rightarrow16\pi$ sq.units.
The required area =$16\pi-\large\frac{4}{3}$$[4\pi+\sqrt 3]$
$\Rightarrow\large\frac{4}{3}$$[4\times 3\pi-4\pi-\sqrt 3]$
$\Rightarrow\large\frac{4}{3}$$[8\pi-\sqrt 3]sq.units.$
Hence the required area=$\large\frac{4}{3}$$[8\pi-\sqrt 3]sq.units.$
answered Sep 25, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai

Related questions