Using integration, find the area of the region bounded by the parabola y^{2}= 4x and the circle 4x^{2}+4y^{2}= 9.

 

 

 

 
 
 
 
 

Answers (1)


for the circle 4x^{2}+4y^{2}= 9,\; we\; have
x^{2}+y^{2}= \left ( \frac{3}{2} \right )^{2}
\therefore centre\; is\; at\left ( 0,0 \right )\S \; r= \frac{3}{2}
Also vertex of the parabola y^{2}= 4x\; is\; at\left ( 0,0 \right ). solving the equations of curves, we get:
4x^{2}+16x-9= 0\Rightarrow \left ( 2x-1 \right )\left ( 2x+9 \right )= 0
\therefore x= \frac{1}{2}\left [ x\not\equiv \frac{-9}{2} \right ]
As the circle and parabola, both are symmetrical about the x-axis
\therefore Required\; area= 2\left [ \int_{0}^{\frac{1}{2}} 2\sqrt{x}\, dx +\int_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}dx}\right ]
\Rightarrow 2\left [ \frac{4}{3}\left [ x\sqrt{x} \right ] \right ]^{\frac{1}{2}}_{0}+\left [ \frac{x}{2}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}+\frac{9}{8}\sin^{-1}\frac{2x}{3}\right ]^{\frac{3}{2}}_{\frac{1}{2}}
\Rightarrow 2\left [ \frac{4}{3} \left [ \frac{1}{2\sqrt{2}}-0 \right ]+\left [ \frac{3}{4} \sqrt{0}+\frac{9}{8}\sin^{-1}1\right ]-\left [ \frac{1}{4}\frac{\sqrt{9-1}}{\sqrt{4}} +\frac{9}{8}\sin^{-1}\frac{1}{3}\right ]\right ]
\Rightarrow 2\left [ \frac{2\sqrt{2}}{6}+\frac{9}{8} \times \frac{\pi }{2}-\frac{\sqrt{2}}{4}-\frac{9}{8}\sin^{-1}\frac{1}{3}\right ]
\Rightarrow \left [ \frac{\sqrt{2}}{6}+\frac{9}{4} \cos^{-1}\frac{1}{3}\right ]sq\cdot units

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