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Using integration, find the area of the region in the first qu

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Using integration, find the area of the region in the first quadrant enclosed by the $x - a x i s$ , the line $y = x$ and the circle $x^{2} + y^{2} = 32$ .

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Eleanor Discussed

Using integration, find the area of the region in the first quadrant enclosed by the

x - a x i s

, the line

y = x

and the circle

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

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Equation of the circle given is

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

Radius of the given circle is

= 32 = 42

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

\Rightarrow y = 32 - x^{2}

(4, 4)

is the point of Intersection of the circle and the line

y = x

This can be calculated by solving the equations

y = x

and

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

simultaneously

The required area is the area of region

O A B

A = (42, 0), B = (4, 4)

and

O = (0, 0)

Area

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

= (\frac{x ^{2}}{2})_{0}^{4} + (\frac{x 32 - x ^{2}}{2} + \frac{32}{2} sin^{- 1} (\frac{x}{4 2}))_{4}^{42}

= 8 + 0 - 8 + 16 \frac{π}{2} - 16 \frac{π}{4}

= 4 π

This is the required answer.

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Question Text	Using integration, find the area of the region in the first quadrant enclosed by the $x - a x i s$ , the line $y = x$ and the circle $x^{2} + y^{2} = 32$ .
Answer Type	Text solution:1
Upvotes	150