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Change the order of the integration of

$\int^1_0 \int^{1+\sqrt{1-y^2}}_{-\sqrt{2y – y^2}} f (x,y) dx \ dy$

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Solution:

Let $\quad I=\int_{0}^{1} \int_{-\sqrt{2 y-y^{2}}}^{1+\sqrt{1-y^{2}}} f(x, y) d x d y$

$\begin{aligned} \text { Region of integration : }-\sqrt{2 y-y^{2}} & \leq x \leq 1+\sqrt{1-y^{2}} \\ 0 & \leq y \leq 1 \end{aligned}$

curves: (i) $x=-\sqrt{2 y-y^{2}} \Rightarrow x^{2}+y^{2}=2 y \Rightarrow x^{2}+(y-1)^{2}=1$

circle with centre $(0,1)$ and radius 1.

(ii) $x=1+\sqrt{1-y^{2}}=\gt\quad(x-1)^{2}+y^{2}=1$

Circle with centre $(1,0)$ and radius $1 .$

(iii) $y=0$ line $i .e$ equation of $x-$axis.

(iv) $y=1$ line parallel to $x-$axis.

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Divide the region R into $R 1$ and $R 2$

$\therefore R=R 1 \cup R 2$

After changing the order of integration,

$\begin{aligned} \text { For region } \mathrm{R} 1 : & 0 \leq y \leq 1-\sqrt{1-x^{2}} \\ & 0 \leq x \leq 1 \end{aligned}$

$For rigion R {2} $: $0 \leq y \leq \sqrt{1-(x-1)^{2}}$ ,$1 \leq x \leq 2$

As the region is divided in two parts the integration will be the union of the two region limits.

$\mathrm{I}=\int_{0}^{1} \int_{0}^{1-\sqrt{1-x^{2}}} f(x, y) d y d x+\int_{1}^{2} \int_{0}^{\sqrt{1-(x-1)^{2}}} f(x, y) d y d x$

This is the integration after changing order from dx dy to dy dx of given integration region.

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