Question

Evaluate `int int(2xy^5)/sqrt(x^2y^2-y^4+1)dxdy`, where R is triangle whose vertices are (0,0),(1,1),(0,1).

Answer 1

let I =`int int(2xy^5)/sqrt(x^2y^2-y^4+1)dxdy`
Region of integration : Triangle whose vertices are (0,0),(1,1),(0,1)

The equation of lines from diagram are : y=1,x=y
0 ≤𝒙≤𝒚
𝟎≤𝒚≤𝟏

`therefore "I"=int_0^1int_0^y(2xy^5)/sqrt(x^2y^2-y^4+1)dxdy`

`=int_0^1int_0^y(2y^5x)/sqrt((1-y^4)+x^2y^2)dxdy`

`=int_0^1int_0^y(2y^5x)/sqrt((1-y^4)/y^2+x^2)1/ydxdy`

`=int_0^1 2y^4[sqrt((1-y^4)/y^2+x^2)]_0^ydy`

`=int_0^1 2y^4[1/y-sqrt(1-y^4)/y]dy`

`=2int_0^1 [y^3-sqrt(1-y^4)y^3]dy`

`=2[y^4/4+1/4(1-y^4)^(3/2)/(3/2)]_0^1`

`=2[1/4-1/4. 2/3]`

`therefore  "I" = 1/6`