We are tasked to evaluate the double integral:

$\iint_R y^2 x \, dA,$ where the rectangle $R$ is defined by: $R = \{(x, y) : -3 \leq x \leq 2, \, 0 \leq y \leq 1\}.$

Solution:

The limits for $x$ are $-3 \leq x \leq 2$, and for $y$, they are $0 \leq y \leq 1$. The integral can be rewritten as: $\int_{y=0}^{1} \int_{x=-3}^{2} y^2 x \, dx \, dy.$

Step 1: Inner integral with respect to $x$

For a fixed $y$, compute: $\int_{x=-3}^{2} y^2 x \, dx.$ Since $y^2$ is treated as a constant in this integral: $\int y^2 x \, dx = y^2 \cdot \frac{x^2}{2} + C.$ Apply the limits $x = -3$ to $x = 2$: $\left[ y^2 \cdot \frac{x^2}{2} \right]_{-3}^{2} = y^2 \cdot \frac{2^2}{2} - y^2 \cdot \frac{(-3)^2}{2}.$ Simplify: $y^2 \cdot \frac{2^2}{2} = y^2 \cdot 2, \quad y^2 \cdot \frac{(-3)^2}{2} = y^2 \cdot \frac{9}{2}.$ Thus: $y^2 \cdot 2 - y^2 \cdot \frac{9}{2} = y^2 \left(2 - \frac{9}{2}\right) = y^2 \left(-\frac{5}{2}\right) = -\frac{5}{2} y^2.$

Step 2: Outer integral with respect to $y$

Now, integrate the result over $y$: $\int_{y=0}^{1} -\frac{5}{2} y^2 \, dy.$ Factor out the constant $-\frac{5}{2}$: $-\frac{5}{2} \int_{y=0}^{1} y^2 \, dy.$ The integral of $y^2$ is: $\int y^2 \, dy = \frac{y^3}{3}.$ Apply the limits $y = 0$ to $y = 1$: $\left[\frac{y^3}{3}\right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}.$ Thus: $-\frac{5}{2} \cdot \frac{1}{3} = -\frac{5}{6}.$

Final Answer:

$\boxed{-\frac{5}{6}}$

Let me know if you'd like further clarifications!

Questions to Explore Further:

1. What is the general procedure for switching the order of integration in double integrals?
2. How would the result change if the limits for $x$ or $y$ were different?
3. Can you explain the geometric interpretation of this integral?
4. What happens if the function to integrate, $y^2x$, has a discontinuity within the region $R$?
5. How can we evaluate the integral using polar coordinates if the region $R$ is rectangular?

Tip:

Always verify the bounds of integration to ensure correctness, especially when integrating iteratively.