Calculus

Application of Integrals

Easy

The portion of the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , that lies in the first quadrant is

$\pi ab$

$\frac{\pi ab}{2}$

$\frac{\pi ab}{4}$

$2\pi ab$

✅ Correct Option: 3

Total area of ellipse $= \pi a b$ . By symmetry, the ellipse is divided into four equal regions by the coordinate axes. Therefore the area in the first quadrant $= \frac{\pi a b}{4}$ .

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The area (in square units) bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$ in first quadrant is equal to:

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The area enclosed between $y^2 = 4x$ , $x = 1$ , $x = 4$ in first quadrant is :

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The area bounded by $2x+y=8, y=a$ and $y=a+2$ with y-axis is 5 sq.units where $a>0$ , then the value of $a$ is:

The portion of the area enclosed by the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1a2x2​+b2y2​=1, that lies in the first quadrant is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The portion of the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , that lies in the first quadrant is