# Transformations

• April 28th 2010, 11:47 AM
Transformations
I am just a little confused by the wording of this problem and I am hoping that someone might have a better idea of how to go about solving this.

Suppose that $\int\int_{D}f(x,y) dA= 3$ where D is the disk $x^2 + y^2 \leq 1$. Now suppose that E is the disk $x^2 + y^2 \leq 4$ and $g(x,y) = 3f(\frac{x} {2}, \frac{y} {2})$. What is the value of $\int\int_{E}g(x,y)dA$?

I am still trying to figure out how to do transformations, so the first thing I did was define two variables and solved for x and y:

$u = \frac{x} {2} and v = \frac{y} {2} \rightarrow 2u = x and 2v = y$

The Jacobian that I found from this ended up being $4dudv$ but I am not sure where to go from here but this is what I attempted:

$x^2 + y^2 \leq 4 \rightarrow (2u)^2 + (2v)^2 \leq 4 \rightarrow u^2 + v^2 \leq 1 \rightarrow u = + or - \sqrt{1^2-v^2}$

This is about as far as I am right now. I am uneasy about whether this is even right so far and how to set up this integral. Thanks for the help ahead of time, I appreciate it.
• April 28th 2010, 12:05 PM
$\iint_E g(x,y)\,\mathrm{d}A = \iint_E 3f(\frac x2,\frac y2)\,\mathrm{d}A = 3 \iint _D f(x,y)\cdot 4\,\mathrm{d}A = 12 \iint_D f(x,y)\,\mathrm{d}A$
Thanks! Let me just make sure that I understand the final product. So because $g(x,y) = 3f(\frac{x} {2},\frac{y} {2})$ and the Jacobian result was 4, the two together yielded $12\int\int_{D} f(x,y)dA$ and because $\int\int_{D} f(x,y)dA = 3$, the end result is 36 then. There would be no need to actually do the integration then right?