Question

Find the area of the region lying to the right of x=y^2-1 and to the left of x=7-y^2

Answer 1

First find intersection by equating values of y^2,
x+1=7-x
Solve for x
x=3
consequently
y^2=7-x=4 =>

`y=\pm2`

The intersection points of the two curves are therefore (3,2) and (3,-2).

The area enclosed by the two curves is therefore obtained by integration within appropriate limits.



`Area=\int_(-2)^(+2)\int_(y^2-1)^(7-y^2)1dxdy`

`=\int_(-2)^(+2) [x]_(y^2-1)^(7-y^2)dy`

`=\int_(-2)^(+2) (8-2y^2)dy`

`=[8y-(2y^3)/(3)]_(-2)^(2)`

`=8(2-(-2))-(32)/(3)`

`=32-(32)/(3)`

`=(64)/(3)`

Answer: Area of between the two curves is
`(64)/(3)`