Double integration using transformation.

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Question 3. A transformation is defined for points (x,y) of not lying on the line L whose equation is x+y=0 by the formulae

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

Solution.

(a) We first find the Jacobian of the transformation

, and we need to remember to take modulus of this value.

(Note. )

(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).

(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.

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?