# Evaluate the double integral over the rectangular region R Double integral with region R (xy)/sqrt ((x^2+y^2+1))  dA ; R={(x,y): 0<=x<=1, 0<=y<=1}

You need to evaluate the iven double integral, over the region `R = {(x,y)/ x in [0,1]; y in [0,1]}`  such that:

`int int_R (xy)/(sqrt(x^2+y^2+1))dA` =`int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx dy`

Converting the double integral into an iterated integral, yields:

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = int_0^1 y*int_0^1 x/(sqrt(x^2+y^2+1)) dx dy`

Notice that `x/(sqrt(x^2+y^2+1)) dx`  represents the derivative of `(sqrt(x^2+y^2+1)), ` considering y as constant such that:

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = int_0^1 y(sqrt(x^2+y^2+1))|_0^1`

Using the fundamental theorem of calculus yields:

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy`

Using the property of linearity of integral yields:

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = int_0^1 ysqrt(y^2+2) dy- int_0^1 ysqrt(y^2+1) dy`

You need to use the following substitution such that:

`t = y^2 + 2 `

Differentiating with respect to t yields:

`dt = 2ydy => ydy = (dt)/2`

You need to change the limits of integration such that:

`y = 0 => t = 2`

`y = 1 => t = 3`

Changing the variable yields:

`int_2^3 sqrt(t) (dt)/2 = (1/2) int_2^3 t^(1/2) dt`

`(1/2) int_2^3 t^(1/2) dt = (1/2) (t^(1/2+1))/(1/2+1)|_2^3`

`(1/2) int_2^3 t^(1/2) dt = (3sqrt3 - 2sqrt2)/3`

You need to solve the definite integral `int_0^1 ysqrt(y^2+1) dy` , hence, you need to come up with the following substitution, such that:

`t =y^2+1 => dt = 2y dy => ydy = (dt)/2`

Changing the limits of integration yields:

`y = 0 => t = 1`

`y = 1 => t = 2`

`(1/2) int_1^3 t^(1/2) dt = (1/2) (t^(1/2+1))/(1/2+1)|_1^2`

`(1/2) int_1^3 t^(1/2) dt = (2sqrt2 - 1)/3`

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = (3sqrt3 - 2sqrt2 - 2sqrt2+ 1)/3`

`int int_R (xy)/(sqrt(x^2+y^2+1))dA = (3sqrt3 - 4sqrt2 + 1)/3`

Hence, evaluating the double integral over the given region R yields  `int int_R (xy)/(sqrt(x^2+y^2+1))dA = (3sqrt3 - 4sqrt2 + 1)/3.`