Jacob ⊕ Jackie ≡ Jacobians Galore

Question

Sketch the the region D bounded by the curves y= 2 / x ,y = 1 / (2x) , y = 2x and y = x/ 2 in the first quadrant of the x y - plane. Let F be the map u = x y and v = y / x from the x y - plane to u v- plane.

A . Find the image of D under F.

B. Let G = F^-1 . Show that | Jac (G ) | = 1 / ( 2|v| )

Γ. Apply the Change of Variables Formula to prove the formula

∫∫_D ƒ (y/x) d x d y = ( 3 / 4) ∫²_1/2 (1/v)ƒ(v) d v

Δ. Apply ( Γ ) to evaluate ∫∫_D [( y e^y/x) / x ] d x d y.

Answer 1

A fun problem,2

I'd like you first try to work the problem using the instructions, then look at the details if necessary.

For a, map each of the curves into (u,v). Note the range of x on each of the paths. This lets you determine the image.

For b and c, we change the integral into the integral of the transformed function times |Jac(G)| over the uv region.

Now, for the details:

I leave it to you to draw the original 4 curves

as (u, v) = (xy,y/x) -

(x, 2/x) maps to (x* 2/x, 2/x/x) = (2, 2/x^2)

(x, 1/(2x)) maps to (x * 1/(2x),1/(2x)/x) = (1/2,1/(2x^2))

(x, 2x) maps to (x*2/x,2x/x) = (2x^2, 2)

(x, x/2) maps to (1/2x^2, 1/2)

y=2/x intersects y = 2x at (1, 2)

y=2/x intersects y = x/2 at (2, 1)

y=1/2x intersects y = 2x at (1/2, 1)

y=1/2x intersects y = x/2 at (1, 1/2)

Thus, (2, 2/x^2), as x ranges from 1 to 2, ranges from (2, 1/2) to (2, 2)

(1/2,1/(2x^2)), as x ranges from 1/2 to 1, ranges from (1/2, 1/2) to (1/2, 2)

(2x^2, 2), as x ranges from 1/2 to 1, ranges from (1/2, 2) to (2, 2)

(1/(2x^2), 1/2), as x ranges from 1 to 2, ranges from (1/2, 1/2) to (2, 1/2)

The image of D under F is a square, [1/2, 2] x [1/2, 2]

B. We calculate the determinant of F, then its inverse.

As (u, v) = (xy, y/x), the Jacobian is the absolute value of the determinant of

y x

-y/x^2 1/x

which has determinant |y/x - -y/x| = |2y/x| = |2v|.

Thus, the inverse has determinant 1/|2v| = 1/(2|v|)

I use I[a, b] for the integral from a to b and E[a, b] for the evaluation from a to b

Then, the integral over D of f(y/x) dx dy =

I[1/2, 2] I[1/2, 2] f(v) |Jac(G)| du dv =

I[1/2, 2] I[1/2, 2] f(v)/(2v) du dv =

I[1/2, 2] u f(v)/(2v) E[1/2, 2] =

I[1/2, 2] (2 - 1/2) f(v)/2v dv =

3/4 I[1/2, 2] f(v)/v dv

Then, the integral of y e^(y/x)/x = (y/x)e^(y/x) over D = as f(v) = ve^v

3/4 I[1/2, 2] ve^v/v dv =

3/4 I[1/2, 2] e^v dv =

3/4 e^v E[1/2, 2] =

3/4 e^2 - 3/4 e^(1/2)