Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y=9x^2$, $y=\frac{x^2}{16}$, and the line $y=1$.

Since both parabolas are symmetric about the y-axis, the total area bounded by the curves is twice the area in the first quadrant.

Let us integrate with respect to $y$. For a given $y$ in the first quadrant, the x-coordinates on the parabolas are:
For $y=\frac{x^2}{16}\Rightarrow x=4\sqrt{y}$ (outer curve)
For $y=9x^2\Rightarrow x=\frac{\sqrt{y}}{3}$ (inner curve)

The required area $A$ is given by:
$A=2{\int }_0^1\left(4\sqrt{y}-\frac{\sqrt{y}}{3}\right)dy$

$A=2{\int }_0^1\left(\frac{11}{3}\sqrt{y}\right)dy$

$A=\frac{22}{3}{\left[\frac{y^{3/2}}{3/2}\right]}_0^1$

$A=\frac{22}{3}\times \frac{2}{3}=\frac{44}{9}$ sq. units

Answer: $\frac{44}{9}$ sq. units