Evaluate the double integral of x^2 + y^2 over the region 0 <= x <= 1, 0 <= y <= 1.

Step-by-step solution

1. 1

   Set up the iterated integral over a rectangle. Constant limits mean the order does not matter, so write $\int_0^1\int_0^1\left(x^2+y^2\right)dy\,dx$.

2. 2

   Split the integrand into two single-variable pieces. $\iint(x^2+y^2)\,dA=\iint x^2\,dA+\iint y^2\,dA$. Each piece separates because the region is a product of intervals.

3. 3

   Evaluate the first piece. $\int_0^1 x^2\,dx\cdot\int_0^1 dy=\frac13\cdot 1=\frac13$.

4. 4

   Use symmetry for the second piece. Swapping the roles of $x$ and $y$ maps the square to itself and turns the second integrand into the first, so $\iint y^2\,dA=\frac13$ as well. Recognising the symmetry saves repeating the computation.

5. 5

   Add the two contributions. The value is $\frac13+\frac13=\frac23\approx 0.6667$.

6. 6

   Numerical check. Adaptive quadrature over the unit square returns $0.66666667$, matching $\frac23$.