Reverse the order of integration in the iterated integral \int_{0}^{1} \int_{\sqrt{y}}^{2} f(x, y) d x d y.
Question 5.2.7: Reverse the order of integration in the iterated integral ∫(...
The region of integration is sketched in Figure 11. This region is divided into two subregions \Omega_{1} and \Omega_{2}. What happens if we integrate first with respect to y ? In \Omega_{1}, 0 \leq y \leq x^{2}. In \Omega_{2}, 0 \leq y \leq 1. Thus
\begin{aligned}\int_{0}^{1} \int_{\sqrt{y}}^{2} f(x, y) d x d y &=\iint_{\Omega} f(x, y) d A=\iint_{\Omega_{1}} f(x, y) d A+\iint_{\Omega_{2}} f(x, y) d A \\&=\int_{0}^{1} \int_{0}^{x^{2}} f(x, y) d y d x+\int_{1}^{2} \int_{0}^{1} f(x, y) d y d x .\end{aligned}
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