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by **kitarzan** » Sun Jul 26, 2009 11:43 pm

Hi, I am having some difficulty with the following problem:

The region R is bounded by the lines y=x, y=x+2, y=2-2x, and y=6-2x. Find the mass of this lamina if the density(x,y)=4x^2+4xy+y^2 g/cm^3.

This problem is supposed to be solved using a substitution to transform the region R into a region G in the uv-plane. I have sketched the region R in the xy-plane and have found the intersections of the boundry functions but I do not know how to find the appropriate functions x=g(u,v) and y=h(u,v) to complete the transformation. Any help would be appreciated.

Thanks!
Kit

by **galactus** » Mon Jul 27, 2009 12:39 am

Let's rewrite the line equations as:

$y-x=0, \;\ y-x=2, \;\ y+2x=2, \;\ y+2x=6$

See now what u and v subs to make?.

Let $u=y-x, \;\ v=y+2x$

Solving these for x and y in terms of u and v gives:

$x=\frac{v-u}{3}, \;\ y=\frac{2u+v}{3}$

Now, find the partials and solve the determinant. I will use d for ${\partial}$ for less typing.

$\begin{vmatrix}\frac{dx}{du}&\frac{dx}{dv}\\ \frac{dy}{du}&\frac{dy}{dv}\end{vmatrix}=\begin{vmatrix}\frac{-1}{3}&\frac{1}{3}\\ \frac{2}{3}&\frac{1}{3}\end{vmatrix}=\frac{-1}{3}$

Note that $4x^{2}+4xy+y^{2}=(2x+y)^{2}=v^{2}$

$\frac{1}{3}\int_{2}^{6}\int_{0}^{2}v^{2}dudv$

Now, can you solve?. I have included a graph of the region pre-transformation and post-transformation

by **kitarzan** » Mon Jul 27, 2009 12:55 am

Thanks! I can't believe I got stumped on that one. Makes perfect sense now.

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