To find the area under the curve $y = x^2$ from $x=0$ to $x=3$, we evaluate the definite integral $\int_{0}^{3} x^2 dx$. Using the power rule of integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we get $[\frac{x^3}{3}]_{0}^{3}$. Substituting the upper limit: $\frac{3^3}{3} = \frac{27}{3} = 9$. Substituting the lower limit: $\frac{0^3}{3} = 0$. The area is $9 - 0 = 9$ sq units. Option 2 is incorrect as it results from a wrong power rule application. Option 3 neglects the division by 3. Option 4 is the area if the function were linear ($y=x$).