MCQ
Choose the correct answer: Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is:
- ✓$\pi$
- B$\frac{\pi}{2}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{4}.$
✓
Answer
Correct option: A.
$\pi$
The equation of circle is $x^2 + y^2 = 4$
we are to find the area of the circle lying between the circle lying between the lines $x = 0$ and $x = 2$ in the first quadrant.
Required area $=\int\limits^2_0\text{y dx}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}\ \ [\therefore$ of $(1)]$
$=\int\limits^2_0\sqrt{(2)^2-\text{x}^2}\text{ dx}$ $=\Big[\frac{\text{x}}{2}\sqrt{(2)^2-\text{x}^2}+\frac{(2)^2}{2}\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{2}}{2}\sqrt{4-\text{4}}+2\sin^{-1}\Big(\frac{\text{2}}{2}\Big)\Big]-[0+2\sin^{-1}0]$
$=\Big[0+2\sin^{-1}(1)-[0+2\times0]\Big]$
$=2\sin^{-1}1=2\times\frac{\pi}{2}=\pi$
we are to find the area of the circle lying between the circle lying between the lines $x = 0$ and $x = 2$ in the first quadrant.
Required area $=\int\limits^2_0\text{y dx}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}\ \ [\therefore$ of $(1)]$
$=\int\limits^2_0\sqrt{(2)^2-\text{x}^2}\text{ dx}$ $=\Big[\frac{\text{x}}{2}\sqrt{(2)^2-\text{x}^2}+\frac{(2)^2}{2}\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\Big]^2_0$
$=\Big[\frac{\text{2}}{2}\sqrt{4-\text{4}}+2\sin^{-1}\Big(\frac{\text{2}}{2}\Big)\Big]-[0+2\sin^{-1}0]$
$=\Big[0+2\sin^{-1}(1)-[0+2\times0]\Big]$
$=2\sin^{-1}1=2\times\frac{\pi}{2}=\pi$
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