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Given a double integral $\iint_{R}f(x,y)dxdy$ continuous on R, find the limits of integration for a given R.

$$R=\{(x,y)|(x-1)^{2}+(y-2)^{2}\leq 1\}$$

I was able to complete one with explicit limits (e.g. $x^2\leq y\leq \sqrt{x}$) but I'm having trouble even imagining how to go about this as if I clear it like this I end up with one expression in terms of $y$ and one in terms of $x$: $$x^2-2x+1 \leq -y^2+4y-3$$

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  • $\begingroup$ This is a unit disk centered at $(1,2)$. $\endgroup$ – Dave May 29 '17 at 19:10
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The boundary of the region is the circle: $(x-1)^2 + (y-2)^2 = 1\implies (y-2)^2 = 1 - (x-1)^2 = 1 - (x^2-2x+1) = -x^2 + 2x\implies y = 2 \pm \sqrt{-x^2+2x}$. Note that $0 \le x \le 2 \implies I = \displaystyle \int_{0}^2 \int_{2-\sqrt{-x^2+2x}}^{2+\sqrt{-x^2+2x}} f(x,y)dydx$.

Similarly, $(x-1)^2 = 1 - (y-2)^2 = 1 - (y^2-4y+4) = -y^2+4y-3\implies x = 1 \pm \sqrt{-y^2+4y-3}$. Note that $1 \le y \le 3 \implies I = \displaystyle \int_{1}^3 \int_{1-\sqrt{-y^2+4y-3}}^{1+\sqrt{-y^2+4y-3}}f(x,y)dxdy$ .

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