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The equation of the circle is $x^2 + y^2 = r^2$, where the radius $r = 8$ cm. So, $x^2 + y^2 = 8^2 = 64$. The equation of the line (scratch) passing through the origin with an angle of $\frac{\pi}{4}$ is $y = x$.
We need to find the point of intersection of the line $y = x$ and the circle $x^2 + y^2 = 64$ in the first quadrant. Substitute $y = x$ into the circle equation: $x^2 + x^2 = 64$ $2x^2 = 64$ $x^2 = 32$ $x = \sqrt{32} = 4\sqrt{2}$ (since we are in the first quadrant, $x > 0$) So, $x = 4\sqrt{2}$ and $y = 4\sqrt{2}$. The point of intersection is $(4\sqrt{2}, 4\sqrt{2})$.
The area of the region enclosed by the x-axis, the scratch ($y=x$), and the circular table top in the first quadrant can be found by integrating the difference between the circle's equation and the line's equation from $x=0$ to $x=4\sqrt{2}$. The equation of the circle in terms of $y$ is $y = \sqrt{64 - x^2}$. The area $A$ is given by: $$A = \int_0^{4\sqrt{2}} (\sqrt{64 - x^2} - x) \, dx$$
We can split the integral into two parts: $$A = \int_0^{4\sqrt{2}} \sqrt{64 - x^2} \, dx - \int_0^{4\sqrt{2}} x \, dx$$ The first integral represents the area of a sector of the circle. We can use the formula for the integral of $\sqrt{a^2 - x^2}$: $$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C$$ In our case, $a = 8$. So, $$\int_0^{4\sqrt{2}} \sqrt{64 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{64 - x^2} + \frac{64}{2} \sin^{-1} \left( \frac{x}{8} \right) \right]_0^{4\sqrt{2}}$$ $$= \left[ \frac{4\sqrt{2}}{2} \sqrt{64 - 32} + 32 \sin^{-1} \left( \frac{4\sqrt{2}}{8} \right) \right] - \left[ 0 + 32 \sin^{-1}(0) \right]$$ $$= 2\sqrt{2} \sqrt{32} + 32 \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = 2\sqrt{2} (4\sqrt{2}) + 32 \left( \frac{\pi}{4} \right) = 16 + 8\pi$$ The second integral is: $$\int_0^{4\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_0^{4\sqrt{2}} = \frac{(4\sqrt{2})^2}{2} - 0 = \frac{32}{2} = 16$$ Therefore, the area $A$ is: $$A = (16 + 8\pi) - 16 = 8\pi$$
The area of the region enclosed by the x-axis, the scratch, and the circular table top in the first quadrant is $8\pi$ square cm.
Final Answer: $8\pi$<\/span>
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