1. **State the problem:** Find the volume of the solid obtained by rotating the region bounded by $y=7x^2$, $x=2$, $x=3$, and $y=0$ about the x-axis. 2. **Formula used:** The volume $V$ of a solid of revolution about the x-axis using the disk method is given by: $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $f(x)$ is the function defining the curve and $[a,b]$ is the interval of $x$. 3. **Identify the function and limits:** Here, $f(x) = 7x^2$, $a=2$, and $b=3$. 4. **Set up the integral:** $$V = \pi \int_2^3 (7x^2)^2 \, dx = \pi \int_2^3 49x^4 \, dx$$ 5. **Integrate:** $$\int 49x^4 \, dx = 49 \cdot \frac{x^5}{5} = \frac{49}{5} x^5$$ 6. **Evaluate the definite integral:** $$V = \pi \left[ \frac{49}{5} x^5 \right]_2^3 = \pi \left( \frac{49}{5} 3^5 - \frac{49}{5} 2^5 \right)$$ 7. **Calculate powers:** $$3^5 = 243, \quad 2^5 = 32$$ 8. **Substitute values:** $$V = \pi \left( \frac{49}{5} \times 243 - \frac{49}{5} \times 32 \right) = \pi \frac{49}{5} (243 - 32)$$ 9. **Simplify inside parentheses:** $$243 - 32 = 211$$ 10. **Final volume:** $$V = \pi \frac{49}{5} \times 211 = \frac{10339}{5} \pi$$ **Answer:** $$\boxed{V = \frac{10339}{5} \pi}$$