Find the area of the region in the 1st quadrant enclosed by the x-axis, the line x=31/2y and the circle - Maths - Application of Integrals

Question

Find the area of the region in the 1st quadrant
enclosed by the x-axis, the line x=31/2y and the circle
X2 + Y2 = 4

Koka Sri Lakshmi Divya Sai · Accepted Answer

$The given equations are y= \frac{x}{\sqrt{3}} - - - - (i) x^{2} + y^{2} = 4 - - - (i i) By solving (i) and (ii) we find that the line and circle meet at (\sqrt{3}, 1) in the first quadrant Draw perpendicular BM to the x-axis ∴ Required area=area of the region OBMO+area of the region BMAB now,the area of the region OBMO = \int_{0}^{\sqrt{3}} y ⅆ x = \int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} ⅆ x = {[\frac{x^{2}}{2 \sqrt{3}}]}_{0}^{\sqrt{3}} = \frac{\sqrt{3}}{2} - - - - (i i i) Area of the region BMAB = \int_{\sqrt{3}}^{2} y ⅆ x = \int_{\sqrt{3}}^{2} \sqrt{4 - x^{2}} ⅆ x = {[\frac{1}{2} x \sqrt{4 - x^{2}} + \frac{1}{2} \times 4 \times \sin^{- 1} \frac{x}{2}]}_{\sqrt{3}}^{2} = (\frac{1}{2} \times 2 \times 0 + \frac{1}{2} \times 4 \times \sin^{- 1} (1)) - (\frac{1}{2} \sqrt{3} (1) + \frac{1}{2} \times 4 \times \sin^{- 1} \frac{\sqrt{3}}{2}) = 2 \sin^{- 1} (1) - \frac{\sqrt{3}}{2} - 2 \sin^{- 1} (\frac{\sqrt{3}}{2}) = π - \frac{\sqrt{3}}{2} - \frac{2 π}{3} = \frac{π}{3} - \frac{\sqrt{3}}{2} - - - - (i v) adding (iii) and (iv) we get \Rightarrow \frac{\sqrt{3}}{2} + \frac{π}{3} - \frac{\sqrt{3}}{2} \Rightarrow \frac{π}{3} So required area= \frac{π}{3}$

19

Find the area of the region in the 1st quadrant enclosed by the x-axis, the line x=31/2y and the circle X2 + Y2 = 4

Find the area of the region in the 1^st quadrant enclosed by the x-axis, the line x=3^1/2y and the circle X² + Y² = 4