To set up the integral of $x^2 y$ over the region in the xy-plane enclosed by $x = 0$, $y = 2$, and $y = x$, we need to:

1. Define the region:
   • The boundary $x = 0$ is the y-axis.
   • The line $y = 2$ is a horizontal line at $y = 2$.
   • The line $y = x$ is a diagonal line with a slope of 1.

The region is a triangle bounded by these curves. Specifically:

• For $x = 0$, the y-values range from $y = 0$ to $y = 2$.
• The line $y = x$ intersects the line $y = 2$ at $x = 2$. So, for $x$ ranging from 0 to 2, the upper boundary of the region is $y = x$, and the lower boundary is $y = 0$.

2. Set up the integral:
   • We can integrate with respect to $y$ first, and then with respect to $x$.
   • For a fixed $x$, $y$ ranges from 0 to $x$ (as $y = x$ is the upper boundary for $y$).
   • The limits for $x$ are from 0 to 2.

Thus, the double integral to evaluate $\int \int_R x^2 y \, dA$ (where $R$ is the region of integration) is:

$\int_0^2 \left( \int_0^x x^2 y \, dy \right) dx$

Sketching the region of integration:

To sketch the region:

• Draw the line $x = 0$ (the y-axis).
• Draw the line $y = 2$ (a horizontal line).
• Draw the line $y = x$ (a diagonal line with slope 1). The area of interest is the triangular region bounded by these lines, with vertices at $(0, 0)$, $(2, 2)$, and $(0, 2)$.

Summary:

The integral is set up as:

$\int_0^2 \left( \int_0^x x^2 y \, dy \right) dx$

Would you like further clarification on how to evaluate this integral or the region’s geometry?

Further Exploration:

1. How do you evaluate the inner integral for $y$?
2. How can you change the order of integration for this region?
3. What is the interpretation of this integral in the context of physics or geometry?
4. What would happen if the region of integration was a square instead of a triangle?
5. How can we apply this concept to find the center of mass of a region?

Tip: Always sketch the region first! Visualizing the boundaries makes it much easier to set up the correct limits for the double integral.