1. **State the problem:** We need to find the area bounded by the curve $y=3x^2$ between $x=0$ and $x=6$. 2. **Set up the integral:** The area under the curve from $x=0$ to $x=6$ is given by the definite integral $$\int_0^6 3x^2 \, dx$$ 3. **Integrate the function:** The integral of $3x^2$ with respect to $x$ is $$\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$$ 4. **Evaluate the definite integral:** Substitute the limits into $x^3$: $$\left. x^3 \right|_0^6 = 6^3 - 0^3 = 216 - 0 = 216$$ 5. **Conclusion:** The area bounded by the curve $y=3x^2$ from $x=0$ to $x=6$ is $216$ square units.