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Find the area of the origin enclosed between the two curves \( (x-6)^2+y^2=36\: and \: x^2+y^2=36\)

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  • If 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 two curves are given by x=a and x=b,by taking common values of y from the equation of the two curves.
Step 1:
Given curves are $(x-6)^2+y^2=36$ and $x^2+y^2=36$
The area of the required region is as shown in the fig.
Step 2:
By solving the two equations,we can find the points of intersection of the two circles.
$\Rightarrow 12x=36$
$\Rightarrow 3$
When $x=3,y=\pm 3\sqrt 3$
Hence the points of intersection are $A(3,3\sqrt 3)$ and $A'(3,-3\sqrt 3)$
Step 3:
Required area=2[area of the region ODCAO]
$\qquad\qquad\;\;\;$=2[area of the region ODAOC+area of the region DCAD]
$\qquad\qquad\;\;\;=2[\int_0^3 ydx+\int_3^6 ydx]$
$\qquad\qquad\;\;\;=2[\int_0^3 \sqrt{36-(x-6)^2}dx+\int_3^3\sqrt{ 36-x^2}]dx$
On integrating we get,
$\qquad\qquad\;\;\;=2\big[\large\frac{1}{2}$$(x-6)\sqrt{36-(x-6)^2}+\large\frac{1}{2}$$\times 36\sin^{-1}\big(\large\frac{x-6}{6}\big)\big]_0^3+\large\frac{1}{2}$$(x)\sqrt{36-x^2}+\big[\large\frac{36}{2}$$\sin^{-1}\big(\large\frac{x}{6}\big)\big]_3^{3\sqrt 3}$
$\qquad\qquad\;\;\;=\big[(x-6)\sqrt{36-x^2}+36\sin^{-1}\big(\large\frac{x-6}{6}\big)\big]_3^{3\sqrt 3}$$+x\sqrt{36-x^2}+36\sin^{-1}\big(\large\frac{x}{6}\big)\big]_3^{3\sqrt 3}$
Step 4:
On applying limits,
$\qquad\qquad\;\;\;=[-3\sqrt{ 27}+36\sin^{-1}\big(\large\frac{-1}{2}\big)]$$-[0+36\sin^{-1}(-1)]+[3\sqrt 3\times 3+36\sin^{-1}\big(\large\frac{\sqrt 3}{2}\big)-$$3\sqrt{27}-36\sin^{-1}\big(\large\frac{1}{2}\big)]$
$\qquad\qquad\;\;\;=-6\sqrt{27}-36\times \large\frac{\pi}{6}$$-36\times (-\large\frac{\pi}{2}\big)$$+9\sqrt 3+36\big(\large\frac{\pi}{3}\big)-$$3\sqrt{27}-36\big(\large\frac{\pi}{6}\big)$
$\Rightarrow 30\pi-18\sqrt 3$
answered Oct 3, 2013 by sreemathi.v
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