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CBSE Class 12 Math 2018 Solved Paper

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Question : 26
Total: 29
Using integration, find the area of the region in the first quadrant enclosed by the x axis, the line y = x and the circle x2+y2 = 32
Solution:  

Given that y = x and x2+y2 = 32
Put y = x in x2+y2 = 32
x2+y2 = 32
⇒ 2x2 = 32
⇒ x = ± 4 = y
Hence, point s of intersection are A (4,4) and B (4,-4) .
Required shaded area
=
4
∫
0
 x dx +
√32
∫
4
 √32−x2 dx (Since Region in the first quadrant)
= (
x2
4
)04 +
[
1
2
(√32−x2)+
1
2
×32sin−1(
x
32
)]√324

= 4π sq . units
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