Practicing Success
Practicing Success
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Area bounded by the parabolas $y^2=4x $ and $x^2 =4y$ is : |
$\frac{16}{3}$ 16 $\frac{8}{3}$ $\frac{32}{3}$ |
$\frac{16}{3}$ |
The correct answer is option (1) → $\frac{16}{3}$
finding points of intersection $x^2=4y⇒x^4=4×4y^2$ $⇒x^4=16×4x$ so $x=0$ or $x=4$ $y=0$ or $y=4$ area = $\int\limits_0^42\sqrt{x}-\frac{x^2}{4}dx=\frac{16}{3}$ sq. units |
