Find the area bounded by the circle $x^{2}+y^{2}=16$ and the line $\sqrt{3} y=x$ in the first quadrant, using integration.

CBSE Class 12 · 2017 · Area Under CurvesLong Answer

Find the area bounded by the circle

x^{2} + y^{2} = 16

and the line

3 y = x

in the first quadrant, using integration.

Answer: $\frac{4 π}{3}$ square units

The circle has radius

4

; the line is

y = \frac{x}{3}

. They meet in the first quadrant where

x^{2} + \frac{x ^{2}}{3} = 16 \Rightarrow x = 23, y = 2

.
The region bounded by the line (below) and the circle (above) in the first quadrant is

A = \int_{0}^{23} \frac{x}{3} d x + \int_{23}^{4} 16 - x^{2} d x .

\int_{0}^{23} \frac{x}{3} d x = 23

, and

\int_{23}^{4} 16 - x^{2} d x = [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{23}^{4} = 4 π - 23 - \frac{8 π}{3} = \frac{4 π}{3} - 23 .

A = 23 + \frac{4 π}{3} - 23 = \frac{4 π}{3} square units.

CBSE Class 12 Board · 2017 · Subjective