The area of region \(x,y): 9x^2+16y^2≤144, x≥0, y≥0\ is- Option 1: 3 Sq.units Option 2: 6 Sq.units Option 3: 9 Sq.units Option 4: 12 Sq.units

Answer: 3 Sq.units. Solution: Dividing by 144 gives x^2/16+y^2/9≤1, an ellipse with a=4 and b=3. The first-quadrant portion is one quarter of it: 14 ab=14(4)(3)=3.

Calculus

Application of Integrals

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core

The area of region $\{(x,y): 9x^2+16y^2\le144, x\ge0, y\ge0\}$ is-

$3\pi$ Sq.units

$6\pi$ Sq.units

$9\pi$ Sq.units

$12\pi$ Sq.units

✅ Correct Option: 1

Dividing by 144 gives $\dfrac{x^2}{16}+\dfrac{y^2}{9}\le1$ , an ellipse with $a=4$ and $b=3$ . The first-quadrant portion is one quarter of it: $\dfrac14\pi ab=\dfrac14\pi(4)(3)=3\pi$ .

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The area of region {(x,y):9x2+16y2≤144,x≥0,y≥0}\{(x,y): 9x^2+16y^2\le144, x\ge0, y\ge0\}{(x,y):9x2+16y2≤144,x≥0,y≥0} is-

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The area of region $\{(x,y): 9x^2+16y^2\le144, x\ge0, y\ge0\}$ is-