Using integration, find the area of the following region :
\left \{ \left ( x,y \right ):x^{2}+y^{2}\leq 16a^{2} \; and\; y^{2}\leq 6ax\right \}

 

 

 

 
 
 
 
 

Answers (1)

we have \left \{ \left ( x,y \right ):x^{2}+y^{2} \leq 16a^{2}\; and\: y^{2}\leq 6ax\right \}
consider x^{2}+y^{2}= 16a^{2}---(i)
              y^{2}= 6ax---(ii)
solving (i) and (ii) , x^{2}+6ax-16a^{2}= 0
\Rightarrow \left ( x-2a \right )\left ( x+8a \right )= 0
As x\neq -8a\; \; \therefore x= 2a
so req. area = 2\left [ \int_{0}^{2a}\sqrt{6a}\sqrt{xdx} +\int_{2a}^{4a}\sqrt{\left ( 4a \right )^{2}-x^{2}dx}\right ]
= 2\left [ \left ( \sqrt{6ax}\frac{x\sqrt{x}}{32} \right )^{2a}_{0} +\left ( \frac{x}{2} \sqrt{\left ( 4a \right )^{2}-x^{2}}+\frac{16a^{2}}{2}\sin^{-1}\frac{x}{4a}\right )^{4a}_{2a}\right ]
\Rightarrow \frac{4\sqrt{6a}}{3}\left ( x\sqrt{x} \right )^{2a}_{0}+\left ( x\sqrt{\left ( 4a \right )^{2}-x^{2}} +16a^{2}\sin^{-1}\frac{x}{4a}\right )^{4a}_{2a}
\Rightarrow \frac{4\sqrt{6a}}{3}\left ( 2a\sqrt{2a} \right )+\left [ 0+16a^{2}\times \frac{\pi }{2} \right ]-\left [ 4a^{2}\sqrt{3}+16a^{2}\times \frac{\pi }{6} \right ]
= \frac{16a^{2}\sqrt{3}}{3}-4a^{2}\sqrt{3}+8a^{2}\pi -8a^{2}\times \frac{\pi }{3}
\Rightarrow =\frac{4a^{2}\sqrt{3}}{3}+\frac{16a^{2}\pi }{3}
\Rightarrow =\frac{4a^{2}}{3}\left ( \sqrt{3}+4\pi \right )sq\cdot units

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