Find the area bounded by the circle with equation $x^{2} + y^{2} = 6 x + 7$ , x=0 and area exterior to parabola $3 y^{2} = 16 x$ .

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Solution

The equation of the circle $x^{2} + y^{2} = 6 x + 7$ can be written as $(x - 3)^{2} + y^{2} = 4^{2}$
Point of intersection of the curves would be :-
$3 x^{2} + 16 x = 18 x + 21$
$(3 x + 7) (x - 3) = 0$
$\Rightarrow x = 3, \frac{- 7}{3}$
$x$ can not be negative
$\Rightarrow x = 3 \Rightarrow y = 4$
Required area = $3 \int 0 (\sqrt{16 - (x - 3)^{2}} d x - \sqrt{\frac{16}{3} x} d x$
$= {⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{x - 3}{2} \sqrt{16 - (x - 3)^{2}} + \frac{16}{2} {sin}^{- 1} (\frac{x - 3}{4}) - 2 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{4}{\sqrt{3}} \times 2 x^{\frac{3}{2}}}{3} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦}_{0}^{- 1} = - \frac{3}{2} \sqrt{7} - 8 {sin}^{- 1} \frac{3}{4} - 16 sq. units$

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Similar questions

x^{2} + y^{2} = 8

x^{2} = 2 y

y = x

y \geq 0

x^{2} + y^{2} = 16

y^{2} = 6 x

x^{2} + y^{2} = 16

y^{2} = 6 x

y^{2} = 16 x

x = 4

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