. Evaluate iint R (x ^ 2 + y ^ 2) * dxdy where R is bounded by x = 0 y = 0 x + y = 1

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.

Learn how to evaluate the double integral of (x^2 + y^2) over the region R bounded by x = 0, y = 0, and x + y = 1. This problem involves setting up and solving a double integral using integration techniques and geometric interpretation. Suitable for advanced college-level students.

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