Mathematics - Area Under Curves Question with Solution | TestHub

The given curves are $y^2=9x$ and $x^2=9y$.

Solving the equations to find the points of intersection:

${\left(\frac{x^2}{9}\right)}^2=9x$

$x^4=729x$

$x(x^3-729)=0$

$x=0$ or $x=9$

The corresponding values of $y$ are $0$ and $9$. The points of intersection are $(0,0)$ and $(9,9)$.

The area of the bounded region is given by:

$A={\int }_0^9\left(\sqrt{9x}-\frac{x^2}{9}\right)dx$

$A={\left[3\cdot \frac{2}{3}{x}^{3/2}-\frac{x^3}{27}\right]}_0^9$

$A=2(9{)}^{3/2}-\frac{9^3}{27}$

$A=2(27)-27=27$

Alternatively, the area bounded by the parabolas $y^2=4ax$ and $x^2=4by$ is $\frac{16ab}{3}$.

Here, $4a=9$ and $4b=9$, so $a=\frac{9}{4}$ and $b=\frac{9}{4}$.

Area $=\frac{16}{3}\left(\frac{9}{4}\right)\left(\frac{9}{4}\right)=\frac{81}{3}=27$

Answer: $27$