The area (in sq. units) of the region $\{(x,y): 3x^2 \leq y \leq |x|\}$ is equal to

The area is needed where $3x^2 \leq y \leq |x|$.

The region is below the line $y = |x|$ (upper boundary) and above the parabola $y = 3x^2$ (lower boundary).

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For the region to exist: $3x^2 \leq |x|$

$3x^2 - |x| \leq 0$

For $x \geq 0$, where $|x| = x$:

$3x^2 \leq x$

$3x^2 - x \leq 0$

$x(3x - 1) \leq 0$

Since $x \geq 0$, this gives $0 \leq x \leq \frac{1}{3}$

For $x < 0$, where $|x| = -x$:

$3x^2 \leq -x$

$3x^2 + x \leq 0$

$x(3x + 1) \leq 0$

Since $x < 0$, this gives $-\frac{1}{3} \leq x < 0$

The region exists for $x \in \left[-\frac{1}{3}, \frac{1}{3}\right]$

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Both curves are symmetric about the y-axis. The area for $x \in [0, \frac{1}{3}]$ can be calculated and multiplied by 2.

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For $x \in [0, \frac{1}{3}]$:

$\text{Area (right side)} = \int_0^{1/3} (|x| - 3x^2) \, dx$

$= \int_0^{1/3} (x - 3x^2) \, dx$

$= \left[\frac{x^2}{2} - x^3\right]_0^{1/3}$

$= \frac{(1/3)^2}{2} - (1/3)^3$

$= \frac{1}{18} - \frac{1}{27}$

$= \frac{3}{54} - \frac{2}{54}$

$= \frac{1}{54}$

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$\text{Total Area} = 2 \times \frac{1}{54}$

$= \frac{2}{54}$

$= \frac{1}{27}$

Therefore, the area of the region is $\frac{1}{27}$ square units.