Question

`int int_R e^(-(xy)/2)dA` where `R`:the region bounded by `y=1/4x,y=2x,y=1/x,y=4/x` Evaluate integrals?

`int int_R e^(-(xy)/2)dA` where `R`:the region bounded by `y=1/4x,y=2x,y=1/x,y=4/x`

Answer 1

`= (ln 8 (e^(3/2) - 1) )/(4 e^2)`

Explanation:

graph{(y-2x)(y-1/4 x)(y-1/x)(y-4/x)=0 [-0.5, 5, -0.5, 3]}

You will see from the drawing that you cannot just write down the limits and integrate here, so I have used the first sub that came to mind:

• `{(u = xy),(v = y/x):} qquad {(x = sqrt(u/v)),(y = sqrt(uv)):}`

`implies { (y = 1/4 x), (y = 2x),(y = 1/x ),(y = 4/x ) :} to {( v = 1/4),( v = 2),(u = 1),(u = 4):}`

Jacobian:

`(del(x,y))/(del(u,v)) =det [(x_u, x_v),(y_u, y_v)]`

`=det [(1/(2sqrt(uv)), - 1/2 sqrt(u/v^3)),(1/2sqrt(v/u), 1/2 sqrt(u/v))] = 1/(2v)`

`:. int int_R e^(-(xy)/2) \ dx dy`

`= int_(1/4)^2 \ int_1^4 qquad 1/(2v) e^(-u/2) \ du \ dv`

`= 1/2 int_(1/4)^2 1/v \ dv * int_1^4 e^(-u/2) du`

`= 1/2[ ln v]_(1/4)^2 * [- 1/2 e^(-u/2) ]_1^4`

`= -1/4 [ ln 2 - ln(1/4)] * [ e^(-2) - e^(- 1/2) ]`

`= (ln 8 (e^(3/2) - 1) )/(4 e^2)`

`[approx 0.245]`

I made this up on the hoof so make sure you're happy with it.