To evaluate the double integral

$\int_{0}^{1} \int_{0}^{2} x^2 y \, dx \, dy,$

we proceed as follows:

Step 1: Evaluate the inner integral

The inner integral is:

$\int_{0}^{2} x^2 y \, dx.$

Since $y$ is treated as a constant with respect to $x$, we can factor $y$ out:

$\int_{0}^{2} x^2 y \, dx = y \int_{0}^{2} x^2 \, dx.$

Now, compute $\int_{0}^{2} x^2 \, dx$:

$\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}.$

Thus, the inner integral becomes:

$\int_{0}^{2} x^2 y \, dx = y \cdot \frac{8}{3} = \frac{8}{3} y.$

Step 2: Evaluate the outer integral

The outer integral is now:

$\int_{0}^{1} \frac{8}{3} y \, dy.$

Factor $\frac{8}{3}$ out:

$\frac{8}{3} \int_{0}^{1} y \, dy.$

Now compute $\int_{0}^{1} y \, dy$:

$\int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$

Thus, the outer integral becomes:

$\frac{8}{3} \cdot \frac{1}{2} = \frac{4}{3}.$

Final Answer:

The value of the double integral is:

$\boxed{\frac{4}{3}}$

Would you like details about any of the steps or related questions? Here are 5 questions to explore further:

1. How do we change the order of integration in double integrals?
2. What are some applications of double integrals in real-world problems?
3. How does the presence of a variable limit affect the computation of a double integral?
4. Can you explain the geometric interpretation of this integral?
5. What techniques can simplify double integrals with complex regions?

Tip: Always check whether the limits of integration are constant or variable; this can influence how you set up the integral.