written 5.0 years ago by | • modified 4.0 years ago |
$\int^2_0 \int^{2+\sqrt{4.y^2}}_{2-\sqrt{4 – y^2}} dxdy$
written 5.0 years ago by | • modified 4.0 years ago |
$\int^2_0 \int^{2+\sqrt{4.y^2}}_{2-\sqrt{4 – y^2}} dxdy$
written 5.0 years ago by |
Solution:
1) Given order and given limits: Given order is: first w.r.t. x and
then w.r.t y i.e., a strip parallel to the x-axis varies from $x=2-\sqrt{4-y^{2}}$
to $x=2-\sqrt{4-y^{2}}$ Y varies from $y=0$ to $y=2$ .
2) Region of integration: $x=2-\sqrt{4-y^{2}}$ is the arc and $x=2+\sqrt{4-y^{2}}$ is the arc of the circle $(x-2)^{2}=4$ with centre at $(2,0)$ and radius $=2$ above the x -axis. $y=0$ is the x -axis and $y=2$ is the line parallel to the x -axis through $A(2,2) .$ The region of integration is the semi-circle OAB above the x-axis. The points of intersection of the circle and the x-axis are $O(0,0)$ and $B(4,0)$ .
3) Change of order of integration: To change the order, consider a strip parallel to the y -axis in the region of integration. On this strip y varies from $y=0$ to $y=\sqrt{4-(x-2)^{2}}$ and then strip moves from $x=0$ to $x=4 .$
$\begin{aligned} I &=\int_{0}^{4} \int_{0}^{\sqrt{4-(x-2)^{2}}} d y d x \\ I &=\int_{0}^{4}[y]_{0}^{\sqrt{4-(x-2)^{2}}} d x \\ I &=\int_{0}^{4} \sqrt{4-(x-2)^{2}} d x \end{aligned}$
$I=\left[\frac{x-2}{2} \sqrt{4-(x-2)^{2}}+2 \sin ^{-1} \frac{x-2}{2}\right]_{0}^{4}$
$I=\left(2 \cdot \frac{\pi}{2}\right)-\left(-2 \cdot \frac{\pi}{2}\right)$
$\therefore I=2 \pi$