The area of the smaller region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and the straight line $3x + 4y = 12$ is:

The ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with $a^2 = 16$ and $b^2 = 9$, giving $a = 4$ and $b = 3$.

The ellipse intersects the x-axis at $(\pm 4, 0)$ and the y-axis at $(0, \pm 3)$.

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The line $3x + 4y = 12$ can be rewritten in intercept form by dividing by 12:

$\frac{x}{4} + \frac{y}{3} = 1$

The line crosses the x-axis at $(4, 0)$ and the y-axis at $(0, 3)$.

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For point $(4, 0)$:

$\frac{4^2}{16} + \frac{0^2}{9} = \frac{16}{16} + 0 = 1$

For point $(0, 3)$:

$\frac{0^2}{16} + \frac{3^2}{9} = 0 + \frac{9}{9} = 1$

Both points lie on the ellipse, so the line connects two points on the ellipse in the first quadrant.

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The line divides the first quadrant portion of the ellipse into two regions: a curved region between the ellipse and the line (smaller region), and a triangular region between the line and the axes.

The smaller region area equals the quarter ellipse area minus the triangle area.

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Total ellipse area $= \pi ab$

$= \pi \times 4 \times 3$

$= 12\pi$

Quarter ellipse area $= \frac{12\pi}{4}$

$= 3\pi$ sq units

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The triangle has vertices at $(0, 0)$, $(4, 0)$, and $(0, 3)$.

Triangle area $= \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times 4 \times 3$

$= 6$ sq units

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Smaller region area $= 3\pi - 6$ sq units