To solve this problem, we need to:

1. Interpret the given integral:
   The problem involves changing the order of integration for the given double integral:

   $\int_0^1 \int_y^1 y \, dx \, dy$

2. Determine the region of integration:

   • The inner integral is over $x$ from $x = y$ to $x = 1$.
   • The outer integral is over $y$ from $y = 0$ to $y = 1$.

   The region is bounded by:

   • $x = y$
   • $x = 1$
   • $y = 0$
   • $y = 1$

   This describes a triangular region in the $xy$-plane. Let's visualize it.

3. Change the order of integration:
   To rewrite the integral with $y$ as the inner variable, analyze the bounds:

   • $y$ ranges from $y = 0$ to $y = x$ (as we solve for $y$ given $x \geq y$).
   • $x$ ranges from $x = 0$ to $x = 1$.

   The new integral becomes:

   $\int_0^1 \int_0^x y \, dy \, dx$

4. Evaluate the new integral:

   • First, compute the inner integral: $\int_0^x y \, dy = \left[\frac{y^2}{2}\right]_0^x = \frac{x^2}{2}.$

   • Substitute into the outer integral: $\int_0^1 \frac{x^2}{2} \, dx = \frac{1}{2} \int_0^1 x^2 \, dx.$

   • Compute the outer integral: $\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}.$

   • Multiply by $\frac{1}{2}$: $\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.$

Final Answer:
The value of the integral is:

$\frac{1}{6}.$

Would you like a detailed step-by-step explanation or any further clarifications?

5 Related Questions:

1. What are some tips for determining the bounds of integration when reversing order?
2. How can we geometrically interpret the change of integration order?
3. How does this technique generalize to non-rectangular regions?
4. Can this method be applied in polar coordinates?
5. What is the physical significance of such integrals in physics or engineering?

Tip:

Always sketch the region of integration to visualize the bounds clearly when reversing the order of integration.