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)

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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