The area bounded by the circles $x^{2} + y^{2} = 1, x^{2} + y^{2} = 4$ in the first Quadrant is

\frac{π}{2}

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\frac{3 π}{4}

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3 π

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\frac{π}{4}

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Solution

The correct option is B $\frac{3 π}{4}$
$x^{2} + y^{2} = 1$ and $x^{2} + y^{2} = 4$ are concentric circles
They form Ring in between them.
Area is given by $π (4 - 1) = 3 π$
In first Quadrant it is divided by $4$
So, Area is given by $\frac{3 π}{4}$

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Similar questions

Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is

A. π

The area of the region bounded by the curve $y = x^{2}$ and $y = {sec}^{- 1} [- {sin}^{2} x]$ , (where [.] denotes the greatest integer function), is

\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

Choose the correct answer in the following.

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines x=0 and x=2

(a) $π$ (b) $\frac{π}{2}$

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