Question

Find `int int \ (x^2+y) \ dA` on region bounded by the curves `x=y^2` and `y=x^2`?

Answer 1

`int int_D \ (x^2+y) \ dx \ dy = 33/140`

Explanation:

The region `D` is that bounded by the curves `x=y^2` and `y=x^2=>x=sqrt(y)`

If we integrate wrt `x` first then we consider an infinitesimally thin horizontal strip, so we start from the parabola `color(red)(x=y^2)` and end on the parabola `y=x^2` which corresponds to the positive arc `color(blue)(x=sqrt(y))`. And as we vary `y` we go from `y=0` to `y=1`

So then calculating the double integral by integrating wrt `x` before `y` we can write:

`I = int int_D \ (x^2+y) \ dx \ dy`
`\ \ = int_0^1 int_(y^2)^(sqrt(y)) \ (x^2+y) \ dx \ dy`
`\ \ = int_0^1 {int_(y^2)^(sqrt(y)) \ (x^2+y) \ dx } \ dy`
`\ \ = int_0^1 [x^3/3+xy]_(x=y^2)^(x=sqrt(y)) \ dy`
`\ \ = int_0^1 (y^(3/2)/3+y^(3/2))- (y^6/3+y^3) \ dy`
`\ \ = int_0^1 4/3y^(3/2)- 1/3y^6-y^3 \ dy`
`\ \ = [4/3y^(5/2)/(5/2)- 1/3y^7/7-y^4/4 ]_0^1`
`\ \ = (8/15-1/21-1/4) - 0`
`\ \ = 33/140`

We could also reverse the order of integration.

If we integrate wrt `y` first then we consider an infinitesimally thin vertical strip, so we start from the parabola `color(blue)(y=x^2)` and end on the parabola `color(red)(y^2=x)` which corresponds to `color(red)(y=sqrt(x))`. And as we vary `x` we go from `y=0` to `y=1`

So then calculating the double integral by integrating wrt `x` before `y` we can write:

`I = int int_D \ (x^2+y) \ dy \ dx`
`\ \ = int_0^1 int_(x^2)^(sqrt(x)) \ (x^2+y) \ dy \ dx`
`\ \ = int_0^1 {int_(x^2)^(sqrt(x)) \ (x^2+y) \ dy } \ dx`
`\ \ = int_0^1 [x^2y+y^2/2]_(y=x^2)^(y=sqrt(x)) \ dx`
`\ \ = int_0^1 (x^(5/2)+x/2)-(x^4+x^4/2) \ dx`
`\ \ = int_0^1 x^(5/2)+x/2-3/2x^4 \ dx`
`\ \ = [x^(7/2)/(7/2)+x^2/4-3/10x^5]_0^1`
`\ \ = (2/7+1/4-3/10) - 0`
`\ \ = 33/140`