The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

The given equations of the parabolas are $4y^2 = 9x$ and $3x^2 = 16y$.

From the second equation, we have $y = \dfrac{3x^2}{16}$. Substituting this into the first equation:

$4\left(\dfrac{3x^2}{16}\right)^2 = 9x$

$\dfrac{9x^4}{64} = 9x$

$x^4 - 64x = 0$

$x(x^3 - 64) = 0$

The real roots are $x = 0$ and $x = 4$.

The required area $A$ is the integral of the upper curve minus the lower curve from $x = 0$ to $x = 4$:

$A = \int_{0}^{4} \left( \dfrac{3}{2}\sqrt{x} - \dfrac{3x^2}{16} \right) dx$

$A = \left[ x^{3/2} - \dfrac{x^3}{16} \right]_{0}^{4}$

$A = \left( 4^{3/2} - \dfrac{4^3}{16} \right) - 0$

$A = 8 - 4 = 4$ sq. units.

Answer: 4 sq. units