To evaluate the double integral

$\iint_R (x^2 + y^2) \, dx \, dy$

where $R$ is the region bounded by $x = 0$, $y = 0$, and $x + y = 1$, we first describe the region $R$ in terms of inequalities. The region $R$ is a right triangle with vertices at $(0,0)$, $(1,0)$, and $(0,1)$.

We can set up the integral with respect to $x$ and $y$:

1. For a given $y$, $x$ ranges from $0$ to $1 - y$.
2. $y$ ranges from $0$ to $1$.

Thus, the integral can be written as:

$\iint_R (x^2 + y^2) \, dx \, dy = \int_{0}^{1} \int_{0}^{1-y} (x^2 + y^2) \, dx \, dy$

We will first evaluate the inner integral with respect to $x$:

$\int_{0}^{1-y} (x^2 + y^2) \, dx$

Separate the integrals:

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

First integral:

$\int_{0}^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3}$

Second integral:

$\int_{0}^{1-y} y^2 \, dx = y^2 \int_{0}^{1-y} 1 \, dx = y^2 [x]_0^{1-y} = y^2 (1-y)$

Combining these results:

$\int_{0}^{1-y} (x^2 + y^2) \, dx = \frac{(1-y)^3}{3} + y^2 (1-y)$

Simplify:

$= \frac{(1-y)^3}{3} + y^2 - y^3$

Now, integrate this expression with respect to $y$:

$\int_{0}^{1} \left( \frac{(1-y)^3}{3} + y^2 - y^3 \right) \, dy$

Separate the integrals:

$\frac{1}{3} \int_{0}^{1} (1-y)^3 \, dy + \int_{0}^{1} y^2 \, dy - \int_{0}^{1} y^3 \, dy$

Evaluate each integral:

1. $\frac{1}{3} \int_{0}^{1} (1-y)^3 \, dy$

   Let $u = 1 - y$, $du = -dy$, and change the limits of integration accordingly:

   When $y = 0$, $u = 1$ When $y = 1$, $u = 0$

   So,

   $\frac{1}{3} \int_{1}^{0} u^3 (-du) = \frac{1}{3} \int_{0}^{1} u^3 \, du = \frac{1}{3} \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

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

   $\left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3}$

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

   $\left[ \frac{y^4}{4} \right]_0^1 = \frac{1}{4}$

Combine the results:

$\frac{1}{12} + \frac{1}{3} - \frac{1}{4}$

Find a common denominator (12):

$\frac{1}{12} + \frac{4}{12} - \frac{3}{12} = \frac{1 + 4 - 3}{12} = \frac{2}{12} = \frac{1}{6}$

Thus, the value of the integral is:

$\boxed{\frac{1}{6}}$

Would you like more details or have any questions?

Here are five related questions:

1. How do you convert a double integral into polar coordinates?
2. What is the significance of the Jacobian determinant in changing variables?
3. How do you evaluate a triple integral in cylindrical coordinates?
4. What are some common applications of double integrals in physics?
5. How can you use double integrals to find the area of a region?

Tip: When setting up double integrals, always sketch the region of integration to understand the limits of integration better.