Double integration using transformation.

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Question 3. A transformation is defined for points (x,y) of $\displaystyle R^2$ not lying on the line L whose equation is x+y=0 by the formulae $\displaystyle u=x+y, v=\frac{y}{x+y}.$

Find the Jacobian of the transformation, sketch the contours u=1 and u=1/2 and find the inverse transformation. What is the transformation of the triangle, T, bounded by the x- and y-axes and the line x+y=1? Hence, evaluate the double integral

$\displaystyle \int\int_Te^{\frac{y}{(x+y)}}dxdy.$

Solution.

(a) We first find the Jacobian of the transformation

$\displaystyle J=\frac{\partial(u,v)}{\partial(x,y)}=det(\frac{\p artial(u,v)}{\partial(x,y)})=\frac{x}{(x+y)^2}+\fr ac{y}{(x+y)^2}=\frac{1}{x+y}$, and we need to remember to take modulus of this value.

(Note. $\displaystyle \frac{\partial(u)}{\partial(x)}=1, \frac{\partial(u)}{\partial(y)}=1, \frac{\partial(v)}{\partial(x)}=-\frac{y}{(x+y)^2}, \frac{\partial(v)}{\partial(y)}=\frac{x}{(x+y)^2}$)

(b) sketch the contours u=1 and v=1/2 is easy in the u,v-plane.

(c) find the inverse transformation from f(u,v) to f(x,y).

$\displaystyle u=x+y, x=u-y$

$\displaystyle v=\frac{y}{(u-y+y)}=\frac{y}{u}, y=uv$

$\displaystyle x=u-y=u-uv$

(d) find the transformation of the triangle T in the u,v-coordinates.

In the x,y-coordinates we have: x=0, y=0, x+y=1

Therefore

x+y=1 becomes u=1

x=o means u-uv=0 and u=uv therefore by cancellation v=1

y=0 means uv=0 and u=0, v=0

The resulting area is the rectangle bounded by these four lines.

(e) evaluate the double integral

Bounds of integration (from (d)) are 0 to 1 for u and 0 to 1 for v. The Jacobian of transformation is found in (a), and we note that in the area of integration x+y=u>0 for all u.

$\displaystyle \int\int_Te^{\frac{y}{(x+y)}}dxdy=\int_0^1\int_0^1 e^{v}dudv*u=\int_0^1udu\int_0^1e^vdv=\frac{(e-1)}{2}$

Question about the question. Why in the initial conditions we needed to specify that points (x,y) are NOT lying on the line L whose equation is x+y=0?