Question

solve it guys plz..fast

Answer 1

intersection of x= √3y and X^2 + y^2 = 4

is 3y^2 + y^2 =4

y^2= 1

y= +-1

in 1st quadrant we have to find

so take y= 1

x= √3

Area of region= area of right angled + integration from √3 to 2 , √ 4- x^2 dx

= 1/2× √3× 1 + ( x/2 √4 -x^2 + 4/2 sin^-1 x/2)

= √3/2 + ( 2/2 √ 4 - 4 + 2 sin^-1 2/2 - ( √3/2 √4 -3 + 2 sin^-1 √3/2)

= √3/2 + ( 2 pie/2 - √3/2 - 2 pie/3)

= √3/2 + pie -√3/2 - 2pie/3

= pie/3