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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Choose the correct answer in the the area of the circle \(x^2 + y^2 = 16\) exterior to the parabola \(y^2 = 6x\) is \[\begin{array} (A) \frac{4}{3} (4\pi - \sqrt 3) \quad &(B) \frac{4}{3} (4\pi + \sqrt 3) \quad & (C) \frac{4}{3} (8\pi - \sqrt 3) \quad & (D) \frac{4}{3} (8\pi + \sqrt 3) \end{array}\]

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Toolbox:
  • 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
  • $A=\int _a^b[f(x)-g(x)]dx$
The required area is area of the circle $x^2+y^2=16$ which is extension to the parabola $y^2=6x$
This is shown in the fig:
To find the points of intersection,let us solve the given equation.
substituing for $y^2$ in the equation of the circle we get,
$\;\;\;x^2+6x=16$
$\Rightarrow x^2+6x-16=0$
on factorising we get ,
$\;\;\;(x+8)(x-2)=0$
$\Rightarrow x=-8 \;and\; x=2.$
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
$A=2\begin{bmatrix}\sqrt 6\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_0^2+\frac{x}{2}\begin{bmatrix}\sqrt {16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4}\end{bmatrix}_2^4$
on appliying limits we get,
$A=\frac{4\sqrt 6}{3}(2\sqrt 2)+2\begin{bmatrix}0+\frac{16}{2}\sin^{_1}(1)-\sqrt {16-4}+\frac{16}{2}\sin^{-1}\frac{1}{2}\end{bmatrix}$
$A=\frac{4\sqrt 6}{3}(2\sqrt 2)+2\begin{bmatrix}\frac{8\pi}{2}-4\sqrt 3+8\big(\frac{\pi}{6}\big)\end{bmatrix}$
on simplifying we get,
$A=\frac{4}{3}[4\sqrt 3+6\pi-3\sqrt 3-2\pi]$
$\;\;\;=\frac{4}{3}[\sqrt 3+4\pi]$
But we require the area outer to this area .
Hence by subtracting from the area of circle,
Area of the circle=$\pi r^2$
$\;\;\;\;\;\;\;\;\;\qquad \quad\;\;=\pi(4)^2$
$\;\;\;\;\;\;\;\;\;\qquad \quad\;\;=16\pi$ sq.units.
The required area =$16\pi-\frac{4}{3}[4\pi+\sqrt 3]$
$\;\;\;\;\;\;\;\;\;\qquad \quad\;\;=\frac{4}{3}[4\times 3\pi-4\pi-\sqrt 3]$
$\;\;\;\;\;\;\;\;\;\qquad \quad\;\;=\frac{4}{3}[8\pi-\sqrt 3]sq.units.$
Hence the required area=$\frac{4}{3}[8\pi-\sqrt 3]sq.units.$
The correct answer is C
answered Jan 23, 2013 by sreemathi.v
 

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