CBSE Class 12 · 2018 · Area Under CurvesLong Answer

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line

y = x

and the circle

x^{2} + y^{2} = 32

Answer: $4 π$ square units

Solution

The line

y = x

meets the circle (radius

42

) where

2 x^{2} = 32 \Rightarrow x = 4, y = 4

.
The region runs from the

x

-axis up to the line

y = x

for

0 \leq x \leq 4

, then up to the circle for

4 \leq x \leq 42

A = \int_{0}^{4} x d x + \int_{4}^{42} 32 - x^{2} d x .

\int_{0}^{4} x d x = 8

, and

\int_{4}^{42} 32 - x^{2} d x = [\frac{x}{2} 32 - x^{2} + 16 sin^{- 1} \frac{x}{4 2}]_{4}^{42} = 8 π - (8 + 4 π) = 4 π - 8.

A = 8 + 4 π - 8 = 4 π square units.

CBSE Class 12 Board · 2018 · Subjective