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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Draw a rough sketch of the region {(x,y):$y^2\leq 6ax $ and $x^2+y^2\leq 16a^2$}.Also find the area of the region sketched using method of integration.

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  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\Large\frac{x}{a}\big)$$+c$
Step 1:
Given:{(x,y):$y^2\leq 6ax $ and $x^2+y^2\leq 16a^2$}
Consider $x^2+y^2\leq 16a^2$ or $x^2+y^2=16a^2$
Clearly this curve represents a circle with centre (0,0) and radius $4a$
Consider $y^2\leq 6ax$ or $y^2=6ax$
Clearly this represents a parabola whose vertex is (0,0) and open rightwards.
The region whose area is required is the shaded portion shown in the fig.
Step 2:
Let us find the point of intersection between the two curves.
$y^2=16a^2-x^2$ and $y^2=16ax$
$16a^2-x^2=6ax\Rightarrow y^2=6ax$
$16a^2-x^2=6ax\Rightarrow x^2+6ax-16a^2=0$
On factorizing we get,
$\Rightarrow x=-8a$ and $2a$
Since $x=-8a\Rightarrow y^2=6a(-8a)=$ -ve.
This is inadmirrable,since the value of $y$ are imaginary.
Hence $x=(2a,y)$ and $(2a,-y)$ are the points of intersection.
Step 3:
Now the area of the required region $A=2\int_0^{2a}(y_2+y_1)dx$
Here $y_2=\sqrt{6ax}.$ and $y_1=\sqrt{16a^2-x^2}$
Since the area is enclosed in the I and IV quadrants,
Area=$2\times$ (Area of OAB)
Step 4:
On integrating we get,
$A=2.\sqrt{6a}\begin{bmatrix}\large\frac{x^{\Large\frac{3}{2}}}{\Large\frac{3}{2}}\end{bmatrix}$+$2\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{16a^2-x^2}+\large\frac{16a^2}{2}\normalsize \sin^{-1}\big(\large\frac{x}{4a}\big)\end{bmatrix}_{2a}^{4a}$
On applying limits we get,
$A=2\times \sqrt 2\sqrt 3 a\times\large\frac{2}{3}[\normalsize (2a)^{\Large\frac{3}{2}}-0]+2\begin{bmatrix}\large\frac{4a}{2}\normalsize\sqrt{16a^2-16a^2}+8a^2 \sin^{-1}\big(\large\frac{4a}{4a}\big)\end{bmatrix}$-2$\begin{bmatrix}\large\frac{2a}{2}\normalsize\sqrt{16a^2-4a^2}+\large\frac{16a^2}{2}\normalsize \sin^{-1}\big(\large\frac{x}{4a}\big)\end{bmatrix}$
$\;\;=\large\frac{4\sqrt 2\sqrt 3 a}{3}[\normalsize 2\sqrt 2\times a\sqrt a]+[16a^2\sin^{-1}(1)]-2a(2\sqrt 3a)-16a^2\sin^{-1}\big(\large\frac{1}{2}\big)$
$\;\;=\large\frac{16\sqrt{3}a^2}{3}$$+16a^2\large\frac{\pi}{2}$$-4a^2\sqrt 3-16a^2\large\frac{\pi}{6}$
$\;\;=\large\frac{16\sqrt 3 a^2-12a^2\sqrt 3}{3}+$$8a^2(\pi-\large\frac{\pi}{3})$
$\;\;=\large\frac{4\sqrt 3a^2}{3}+$$8a^2\big(\large\frac{2\pi}{3}\big)=\large\frac{4}{3}$$a^2(\sqrt 3+4\pi)$
Hence the required area is $\large\frac{4}{3}$$a^2(\sqrt 3+4\pi)$
answered May 6, 2013 by sreemathi.v
edited May 7, 2013 by sreemathi.v

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