The area enclosed by the ellipse x^2/16 + y^2/9 = 1 is given by : Option 1: 3 _0^4 √(9 - x^2) dx Option 2: 3/4 _0^4 √(9 - x^2) dx Option 3: 3 _0^4 √(16 - x^2) dx Option 4: 3/4 _0^4 √(16 - x^2) dx

Answer: 3 _0^4 √(16 - x^2) dx. Solution: From x^2/16+y^2/9=1: y=3/4√(16-x^2). Total area =4_0^43/4√(16-x^2) dx=3_0^4√(16-x^2) dx.

Calculus

Application of Integrals

Easy

common

The area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is given by :

$3 \int_0^4 \sqrt{9 - x^2} \, dx$

$\frac{3}{4} \int_0^4 \sqrt{9 - x^2} \, dx$

$3 \int_0^4 \sqrt{16 - x^2} \, dx$

$\frac{3}{4} \int_0^4 \sqrt{16 - x^2} \, dx$

✅ Correct Option: 3

From $\frac{x^2}{16}+\frac{y^2}{9}=1$ : $y=\frac{3}{4}\sqrt{16-x^2}$ .

Total area $=4\int_0^4\frac{3}{4}\sqrt{16-x^2}\,dx=3\int_0^4\sqrt{16-x^2}\,dx$ .

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