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Mathematics Question on applications of integrals

Area enclosed by the curve π[4(x−2)2+y2]=8\pi\left[4\left(x-\sqrt{2}\right)^{2} +y^{2}\right]=8 is

A

π\pi

B

22

C

3 π3\, \pi

D

44

Answer

44

Explanation

Solution

The given curve is
π[4(x−2)2+y2]=8\pi\left[4\left(x-\sqrt{2}\right)^{2} +y^{2}\right]=8
4(x−2)2+y2=8π4\left(x-\sqrt{2}\right)^{2} +y^{2} =\frac{8}{\pi}
(x−2)2+(y2)2=2π\left(x-\sqrt{2}\right)^{2} +\left(\frac{y}{2}\right)^{2} =\frac{2}{\pi}
(x−2)2(2π)2+y2(22π)2=1...(1)\frac{\left(x-\sqrt{2}\right)^{2}}{\left(\sqrt{\frac{2}{\pi}}\right)^{2}}+\frac{y^{2}}{\left(2\sqrt{\frac{2}{\pi }}\right)^{2}}=1\quad\quad...\left(1\right)
This is the equation of the ellipse having centre ( 2\sqrt{ 2},0 ).
Observe the figure of ellipse (1). The centre P is ( 2\sqrt{ 2},0 ). A and B are (2,−2π,0)\left(\sqrt{2},-\sqrt{\frac{2}{\pi},0}\right) and
(2,+2π,0)\left(\sqrt{2},+\sqrt{\frac{2}{\pi},0}\right) respectively.
The required area = 4 ? area of figure PQB
=4×∫22+2π=4\times\int^{ \sqrt{2}+\sqrt{\frac{2}{\pi}}}_{ \sqrt{2}}ydx
=4×∫22+2π8π−4(x−2)2=4\times \int_{\sqrt{2}}^{\sqrt{2}+\sqrt{\frac{2}{\pi }}}\sqrt{\frac{8}{\pi}-4\left(x-\sqrt{2}\right)^{2}} dx
=4×2∫22+2π(2π)2−(x−2)2=4\times2\int^{\sqrt{2}+\sqrt{\frac{2}{\pi }}}_{\sqrt{2}}\sqrt{\left(\sqrt{\frac{2}{\pi}}\right)^{2}-\left(x-\sqrt{2}\right)^{2}} dx
=8[x−222π−(x−2)2+(2/π)2sin−1(x−22/π)]22+2π=8\left[\frac{x-\sqrt{2}}{2}\sqrt{\frac{2}{\pi}-\left(x-\sqrt{2}\right)^{2}}+ \frac{\left(2/\pi\right)}{2}sin^{-1}\left(\frac{x-\sqrt{2}}{\sqrt{2/\pi}}\right)\right]_{_{_{_{_{\sqrt{2}}}}}}^{^{\sqrt{2}+\sqrt{\frac{2}{\pi}}}}
[∵∫a2−x2dx=12xa2−x2+.a22sin−1xa+C]\left[\because \int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}x\sqrt{a^{2}-x^{2}}+.\frac{a^{2}}{2}sin^{-1}\frac{x}{a}+C\right]
=8[{2+2π−222π−(2+2x−2)2+1πsin−1(2+2π−22π)−2−2222π−(2−2)2+1πsin−1(2−22π)}]=8\left[\begin{Bmatrix}\frac{\sqrt{2}+\sqrt{\frac{2}{\pi}}-\sqrt{2}}{2}\sqrt{\frac{2}{\pi}-\left(\sqrt{2}+\sqrt{\frac{2}{x}-\sqrt{2}}\right)^{2}}+\frac{1}{\pi}sin^{-1}\left(\frac{\sqrt{2}+\sqrt{\frac{2}{\pi}-\sqrt{2}}}{\sqrt{\frac{2}{\pi}}}\right)-\frac{\sqrt{2}-\sqrt{2}}{2}2\sqrt{\frac{2}{\pi}-\left(\sqrt{2}-\sqrt{2}\right)^{2}}+\frac{1}{\pi}sin^{-1}\left(\frac{\sqrt{2}-\sqrt{2}}{\sqrt{2\pi}}\right)\end{Bmatrix}\right]
=8[12π(0)+1πsin−1(1)−0−0]=8\left[\frac{1}{\sqrt{2\pi}}\left(0\right)+\frac{1}{\pi}sin^{-1}\left(1\right)-0-0\right]
=8(1π×π2)=8\left(\frac{1}{\pi}\times\frac{\pi}{2}\right)=4 square units.