The area of the smaller region of the circle $x^2 + y^2 = 8$ cut off by the line $x = 2$ is

The circle $x^2 + y^2 = 8$ has center $(0, 0)$ and radius $r = \sqrt{8} = 2\sqrt{2}$.

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Substitute $x = 2$ into the circle equation:

$2^2 + y^2 = 8$

$4 + y^2 = 8$

$y^2 = 4$

$y = \pm 2$

The line intersects the circle at $(2, 2)$ and $(2, -2)$.

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Since the circle's center is at the origin with radius $2\sqrt{2} \approx 2.83$, and the line is at $x = 2$, the region where $x > 2$ is the smaller region.

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The smaller region is a circular segment with area = Sector area - Triangle area.

For the sector area, find the angle subtended by the chord at the center:

Point $(2, 2)$ makes angle $\theta_1 = \tan^{-1}(2/2) = \tan^{-1}(1) = \frac{\pi}{4}$

Point $(2, -2)$ makes angle $\theta_2 = -\frac{\pi}{4}$

Total angle: $\theta = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$

Sector area $= \frac{1}{2}r^2\theta$

$= \frac{1}{2}(8)\left(\frac{\pi}{2}\right)$

$= 2\pi$

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For the triangle with vertices $(0, 0)$, $(2, 2)$, $(2, -2)$:

Base = distance from $(2, 2)$ to $(2, -2)$ = $4$ units

Height = perpendicular distance from origin to line $x = 2$ = $2$ units

Triangle area $= \frac{1}{2} \times 4 \times 2 = 4$

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Area of smaller region = Sector area - Triangle area

$= 2\pi - 4$

$= 2(\pi - 2)$

Therefore, the area of the smaller region is $2(\pi - 2)$ square units.