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Find the area of the region lying above x-axis and included between the circle $x^2 + y^2 = 8x$ and inside of the parabola $y^2 = 4x$

Answers (1)

R Ravindra Pindel

Given: $x^2 + y^2 = 8x$ can be expressed as

$(x - 4)^2 + y^2 = 16\quad -(i)$

Centre is (4,0) & radius 4 and equation of parabola is

$y^2 = 4x\quad -(ii)$

$\therefore I = \int_0^4 y\text{ of the parabola }dx + \int_4^8y\text{ of the circle }dx$

$= \underset{I_1}{\int_0^4 2\sqrt{x}dx} + \underset{I_2}{\int_4^8\sqrt{4^2 - (x-4)^2}dx}$

$I_1 = {\int_0^4 2\sqrt{x}dx}$

$I_1 =2\left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_0$

$I_1 =\frac{4}{3}\left[4^{\frac{3}{2}} - 0 \right ]$

$\Rightarrow \frac{4}{3} (8) = \frac{32}{3}\qquad -(1)$

and $I_2 = {\int_4^8\sqrt{4^2 - (x-4)^2}dx}$

$I_2 = \left [\frac{x-4}{2}\sqrt{4^2 -(x-4)^2 } + \frac{16}{2}\sin^{-1}\frac{x-4}{4}\right ]_4^8$

$I_2 = \left [\frac{x-4}{2}\sqrt{4^2 -(x-4)^2 } + \frac{16}{2}\sin^{-1}\frac{x-4}{4}\right ]_4^8$

$I_2 =\frac{1}{2} \left [\frac{4}{2}(0) + 16\sin^{-1}(1) - (0)\right ]$

$I_2 =\frac{1}{2}\times 16\times \frac{\pi}{2} = 4\pi \qquad -(2)$

$\therefore I = I_1 + I_2$

$= \left (\frac{32}{2} + 4\pi \right )$

$\Rightarrow\frac{4}{3} \left (8 + 3\pi \right ) \ \text{sq units}$

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