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Math Help - Transformations

  1. #1
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    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.
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  2. #2
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    \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
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  3. #3
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    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?

    And again, thank you so much for being so quick.
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  4. #4
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    Actually, I might have gotten that backward. Is it supposed to be 4?

    edit: no, I'm pretty sure that's right. Sorry.
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