Find the area of the region `{(x, y) : y^2 <=6ax and x^2+y^2<=16a^2}` using method of integration .

Answer

N/a

Explanation

We have, `y^(2)=6ax" and " x^(2)+y^(2)=16a^(2)`
`rArr x^(2)+16ax-16a^(2)=0`
`rArr x^(2)+8ax-2ax-16a^(2)=0`
`rArr x(x+8a)-2a(x+8a)=0`
`rArr (x-2a)(x+8a)=0`
`rArr x=2a, -8a`

`:. " Area of required region "=[int_(0)^(2a)sqrt(6ax)dx+int_(2a)^(4a)sqrt((4a)^(2)-x^(2)dx)]`
`=2[int_(0)^(2a)sqrt(6a)x^(1//2)dx+int_(2a)^(4a)sqrt((4a)^(2)-x^(2))dx]`
`=2[sqrt(6a)[(x^(3//2))/(3//2)]_(0)^(2a)+(x/2sqrt((4a)^(2)-x^(2))+(4a)^(2)/2sin^(-1).(x)/(4a))_(2a)^(4a)]`
`=2[sqrt(6a) . 2/3((2a)^(3//2)-0)+(4a)/2 .0+(16^(2))/2 .pi/2-(2a)/2sqrt(16a^(2)-4a^(2))-(16a^(2))/2 .sin^(-1). (2a)/(4a)]`
`=2[sqrt(*6a)2/3 .2sqrt2a^(3//2)+0+4pia^(2)-(2a)/2 .sqrt3a-8a^(2) .pi/6]`
`=2[sqrt12 .4/3a^(2)+4pia^(2)-2sqrt3a^(2)-(4a^(2)pi)/3]`
`=2[(8sqrt3a^(2)+12pia^(2)-6sqrt3a^(2)-4a^(2)pi)/3]`
`=2/3a^(2)[8sqrt3+12pi-6sqrt3a^(2)-4a^(2)pi]`
`=2/3a^(2)[2sqrt3+8pi]=4/3a^(2)[sqrt3+4pi]`