What is int^(1_0) int^(1_0) xy/(sqrt(x^2+y^2+1))?I am mostly confused about how to find the integral of 1/(sqrt(x^2+y^2+1)), thanks!

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to know the following aspect: If you should solve the double integral `int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx dy` , you need to solve the inner integral considering y as constant, but if you should solve the double integral `int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dy dx` , you need to solve the inner integral considering x as constant.

Supposing that you should solve `int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx dy` , you need to start solving inner integral such that:

`int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx = y int_0^1 (x)/(sqrt(x^2+y^2+1)) dx`

You need to substitute t for `x^2+y^2+1`  such that:

`x^2+y^2+1 = t`

If `x = 0 =gt t = y^2 + 1`

If `x = 1 =gt t = y^2 + 2`

Differentiating both sides yields:

`2xdx = dt =gt xdx = (dt)/2`

`y int_(y^2 + 1)^(y^2 + 2) ((dt)/2)/(sqrt(t)) = (y/2)*t^(1/2)/(1/2)|_(y^2 + 1)^(y^2 + 2)`

`y int_(y^2 + 1)^(y^2 + 2) ((dt)/2)/(sqrt(t)) = y*sqrt t|_(y^2 + 1)^(y^2 + 2)`

Substituting `x^2+y^2+1`  for t yields:

`int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx = y*sqrt(x^2+y^2+1)|_0^1`

`int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx = y(sqrt(y^2+2)-sqrt(y^2+1))`

Hence, you need to solve for y the outer integral such that:

`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 y(sqrt(y^2+1)dy`

You should come up with the substitution `y^2+1 = u` , hence, differentiating both sides yields:

`2ydy = du =gt ydy = (du)/2`

Substituting v for `y^2+2`  yields:

`y^2+2 = v =gt 2ydy = dv =gt ydy = (dv)/2 `

If `y=0 =gt u=1 and v = 2`

If`y =1 =gt u= 2 and v = 3`

You should solve the integral `int_0^1 y(sqrt(y^2+2)dy`  such that:

`int_0^1 y(sqrt(y^2+2)dy = (1/2)int_2^3 sqrt v dv`

`(1/2)int_2^3 sqrt v dv = (1/2) (v^(1/2+1))/(1/2+1)|_2^3`

`(1/2)int_2^3 sqrt v dv = (1/3)v*sqrt v|_2^3`

`(1/2)int_2^3 sqrt v dv = (1/3)(3sqrt3 - 2sqrt2)`

You should solve the integral `int_0^1 y(sqrt(y^2+1)dy`  such that:

`int_0^1 y(sqrt(y^2+1)dy= (1/2)int_1^2 sqrtu dv`

`int_0^1 y(sqrt(y^2+1)dy= (1/3)u*sqrt u|_1^2`

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

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

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

Hence, evaluating the double integral `int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx dy`  yields `int_0^1 int_0^1 (xy)/(sqrt(x^2+y^2+1)) dx dy = (1/3)(3sqrt3 - 4sqrt2 + 1).`

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