0
868views
Change the order of integration and evaluate $ \int^1_0 \int^\sqrt{2-x^2}_x \frac{x}{\sqrt{x^2 + y^2}} dx \ dy$
1 Answer
0
3views

Solution:

Let $I=\int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{x d y d x}{\sqrt{x^{2}+y^{2}}}$

Region of integration is: $\quad x \leq y \leq \sqrt{2-x^{2}}\\ 0 \leq x \leq 1$

Curves: $(i) y=x$ line

$\quad(i i) x=0, x=1 \quad$ lines parallel to the y axis

(iii) $y=\sqrt{2-x^{2}} \Rightarrow x^{2}+y^{2}=2$

Circle with centre $(0,0)$ and radius $\sqrt{2}$ .

Intersection of circle and $y=x$ line is $(1,1)$ in $1^{st}$ quadrant.

enter image description here

Divide the region into two parts as shown in fig.

After changing the order of integration:

$\begin{array}{ll}{\text { For one region: }} & {0 \leq x \leq y} \\ & {0 \leq y \leq 1}\end{array}$

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

$\begin{aligned} \therefore \mathbf{I} &=\int_{0}^{1} \int_{0}^{y} \frac{x d x d y}{\sqrt{x^{2}+y^{2}}}+\int_{1}^{\sqrt{2}} \int_{0}^{\sqrt{2-y^{2}}} \frac{x d x d y}{\sqrt{x^{2}+y^{2}}} \\ &=\int_{0}^{1}\left[\sqrt{x^{2}+y^{2}}\right]_{0}^{y} d y+\int_{1}^{\sqrt{2}}\left[\sqrt{x^{2}+y^{2}}\right]_0^{\sqrt{2-y^{2}}} \\ &=\int_{0}^{1}(\sqrt{2} . y-y) d y+\int_{1}^{\sqrt{2}}(\sqrt{2}-1) d y \\ &=(\sqrt{2}-1)\left[\frac{y^{2}}{2}\right]_{0}^{1}+\left[\sqrt{2} y-\frac{y^{2}}{2}\right]_{1}^{\sqrt{2}} \end{aligned}$

$=1-\frac{1}{\sqrt{2}}$

$\therefore I=\frac{\sqrt{2}-1}{\sqrt{2}}$

Please log in to add an answer.