Find the area line in the first quadrant and bounded by the circle x²

1.	Find the area line in the first quadrant and bounded by the circle x²+y²=4 and the lines X=0,X=2
Answer» $\large\underline\purple{\bold{Solution :- }}$ ☆ Equation of given circle is $\tt \: {x}^{2} + {y}^{2} = 4$ $\tt \: {(x - 0)}^{2} + {(y - 0)}^{2} = {(2)}^{2}$ $\tt\implies \:radius \: of \: circle = 2 \: and \: centre \: is \: (0, 0)$ ☆ x = 0 REPRESENTS the y - axis. ☆ x = 2 represents a LINE PARALLEL to y-axis and PASSES through (2, 0). $\red{ \tt \: So \: Required \: area = \: Area \: of \: Shaded \: region}$ $\tt \longrightarrow \: = \: Area \: OAB$ $\tt \longrightarrow \: = \int_0^2 \: y \: dx$ $\tt \longrightarrow \: = \int_0^2 \: \sqrt{ {2}^{2} - {x}^{2} } \: dx$ $\tt \longrightarrow \: = \bigg[ \dfrac{x}{2} \sqrt{ {2}^{2} - {x}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{x}{2} \bigg]_0^2$ $\tt \longrightarrow \: = \bigg[ \dfrac{2}{2} \sqrt{ {2}^{2} - {2}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{2}{2} \bigg] - \bigg[ \dfrac{0}{2} \sqrt{ {2}^{2} - {0}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{0}{2} \bigg]$ $\tt \longrightarrow \: = 0 + 2 {sin}^{ - 1} 1 - 0 - 0$ $\tt \longrightarrow \: = 2 \times \dfrac{\pi}{2}$ $\tt \longrightarrow \: = \pi \: square \: units$

Find the area line in the first quadrant and bounded by the circle x²+y²=4 and the lines X=0,X=2

Answer»

$\large\underline\purple{\bold{Solution :- }}$

☆ Equation of given circle is

$\tt \: {x}^{2} + {y}^{2} = 4$

$\tt \: {(x - 0)}^{2} + {(y - 0)}^{2} = {(2)}^{2}$

$\tt\implies \:radius \: of \: circle = 2 \: and \: centre \: is \: (0, 0)$

☆ x = 0 REPRESENTS the y - axis.

☆ x = 2 represents a LINE PARALLEL to y-axis and PASSES through (2, 0).

$\red{ \tt \: So \: Required \: area = \: Area \: of \: Shaded \: region}$

$\tt \longrightarrow \: = \: Area \: OAB$

$\tt \longrightarrow \: = \int_0^2 \: y \: dx$

$\tt \longrightarrow \: = \int_0^2 \: \sqrt{ {2}^{2} - {x}^{2} } \: dx$

$\tt \longrightarrow \: = \bigg[ \dfrac{x}{2} \sqrt{ {2}^{2} - {x}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{x}{2} \bigg]_0^2$

$\tt \longrightarrow \: = \bigg[ \dfrac{2}{2} \sqrt{ {2}^{2} - {2}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{2}{2} \bigg] - \bigg[ \dfrac{0}{2} \sqrt{ {2}^{2} - {0}^{2} } + \dfrac{ {2}^{2} }{2} {sin}^{ - 1}\dfrac{0}{2} \bigg]$