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

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### 1 Answer

You need to evaluate the following double integral 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`

You need to convert the double integral into an iterated integral, 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 dy`

`int_0^1 y(sqrt(x^2+y^2+1))_0^1 dy = int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy`

Using the property of linearity of integral yields:

`int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy = int_0^1 y(sqrt(y^2+2))dy - int_0^1 ysqrt(y^2+1)) dy`

You need to use the following substitution such that:

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

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) (2t^(3/2))/3|_2^3`

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

`int_0^1 ysqrt(y^2+1)) dy = (1/2) int_1^2 t^(1/2) dt `

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

`int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy = (3^(3/2))/3 - (2^(3/2))/3 - (2^(3/2))/3+ (1^(3/2))/3 `

`int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy = (3sqrt3 - 2sqrt2 - 2sqrt2 + 1)/3`

`int_0^1 y(sqrt(y^2+2) - sqrt(y^2+1)) dy = (3sqrt3 - 4sqrt2 + 1)/3`

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