Evaluate the function f(x, y) = x² + y² over the triangular region with vertices (0, 0), (1, 0), and (0, 1).

Steps to Solve

1. Define the Region of Integration

The triangular region is bounded by the vertices (0, 0), (1, 0), and (0, 1). In the Cartesian plane, this creates a right triangle.

2. Set Up the Double Integral

The function we want to integrate is $f(x, y) = x^2 + y^2$. The limits for $y$ will be from 0 to $1 - x$ for $x$ ranging from 0 to 1:

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

3. Integrate with Respect to $y$ First

Now we compute the inner integral:

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

This can be split into two integrals:

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

The first integral is:

$$ x^2 \cdot y \Big|_{0}^{1-x} = x^2(1-x) $$

The second integral is:

$$ \frac{y^3}{3} \Big|_{0}^{1-x} = \frac{(1-x)^3}{3} $$

So, combining these gives:

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

4. Integrate with Respect to $x$

Now substitute the result from the previous step into the outer integral:

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

This can be evaluated as two separate integrals:

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

5. Evaluate Each Integral

The first integral is:

$$ \int_{0}^{1} x^2(1-x) , dx = \int_{0}^{1} (x^2 - x^3) , dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} $$

The second integral simplifies to:

$$ \int_{0}^{1} (1-x)^3 , dx = \frac{1}{4} $$

Therefore:

$$ \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} $$

6. Combine the Results

Now, sum these results from the two integrals:

$$ \frac{1}{12} + \frac{1}{12} = \frac{1}{6} $$