Question

Evaluate the double integral `int_0^1 int_0^(sqrt(1-x^2)) sqrt(1-y^2) dy dx` by reversing the order of integration?

Answer 1

`int_0^1 int_0^(sqrt(1-x^2)) \ sqrt(1-y^2) \ dy \ dx = 2/3`

Explanation:

We seek:

`I = int_0^1 int_0^(sqrt(1-x^2)) \ sqrt(1-y^2) \ dy \ dx`

If we look at the inner integral first, we have integration limits:

`y` limits are `{ (y = 0), (y = sqrt(1-x^2)) :}`

`x` limits are `{ (x = 0), (x=1) :}`

So the region `D` is as follows:

If we reverse the order of integration the region must obviously remain unaltered, but the integration limits would become:

`x` limits are `{ (x = 0), (x=sqrt(1-y^2)) :}`

`y` limits are `{ (y = 0), (y = 1) :}`

Thus, we can reverse the integration order and the integral remains the same, giving:

`I = int_0^1 int_0^(sqrt(1-y^2)) \ sqrt(1-y^2) \ dx \ dy`

The inner integral is now trivial to integrate:

`I_("Inner") = int_0^(sqrt(1-y^2)) \ sqrt(1-y^2) \ dx`
`\ \ \ \ \ \ \ \ = [(sqrt(1-y^2)) x]_0^(sqrt(1-y^2))`
`\ \ \ \ \ \ \ \ = (sqrt(1-y^2))(sqrt(1-y^2)) - 0`
`\ \ \ \ \ \ \ \ = 1-y^2`

So then returning to our double integral we haver:

`I = int_0^1 I_("Inner") \ dy`
`\ \ = int_0^1 1-y^2 \ dy`
`\ \ = [y-y^3/3]_0^1`
`\ \ = (1-1/3) - (0)`
`\ \ = 2/3`

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Observations:

One might ask why we bother changing the order of integration. Consider the original integral, without any changes:

`I = int_0^1 int_0^(sqrt(1-x^2)) \ sqrt(1-y^2) \ dy \ dx`

Here, the inner integral is:

`I_("Inner") = int_0^(sqrt(1-x^2)) \ sqrt(1-y^2) \ dy`
`\ \ \ \ \ \ \ \ = [(arcsin(x)+xsqrt(1-x^2))/2]_0^(sqrt(1-x^2))`
`\ \ \ \ \ \ \ \ = 1/2( arcsin(sqrt(1-x^2)) + (sqrt(1-x^2)))sqrt(1-(sqrt(1-x^2))^2) )`
`\ \ \ \ \ \ \ \ = 1/2( arcsin(sqrt(1-x^2)) + sqrt(1-x^2) sqrt(1-1+x^2) )`
`\ \ \ \ \ \ \ \ = 1/2( arcsin(sqrt(1-x^2)) + xsqrt(1-x^2) )`

Hopefully the reasoning is already clear, but should we continue we get:

`I = int_0^1 I_("Inner") \ dx`
`\ \ = 1/2 \ int_0^1 arcsin(sqrt(1-x^2)) + xsqrt(1-x^2) \ dx`
`\ \ = 1/2 [xarcsin(sqrt(1-x^2)) - sqrt(1-x^2) -(1-x^2)^(3/2)/3]_0^1`
`\ \ = 1/2 { (0-0-0) - (0-1-1/3)}`
`\ \ = 1/2 { 0 - (-4/3)}`
`\ \ = 2/3`

So although we can perform the integration, significant steps have been omitted to perform the integration, and the problem in this form is significantly more complex.