Question
using integration find the area of the region enclosed between the circle x^2+y^2=16 and the parabola y^2 = 6x
Answers
Steve
The curves intersect where (2,±√12)
Best to integrate on y here, so the area is (using symmetry on x-axis)
2∫[0,√12] √(16-y^2) - y^2/6 dy
= 16arcsin(y/4) - y/9 (y^2 - 9√(16-y^2)) [0,√12]
= 4/√3 + 16π/3
Best to integrate on y here, so the area is (using symmetry on x-axis)
2∫[0,√12] √(16-y^2) - y^2/6 dy
= 16arcsin(y/4) - y/9 (y^2 - 9√(16-y^2)) [0,√12]
= 4/√3 + 16π/3