Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D \frac{x^2}{1+y^2}\,dx\,dy, $$ where $0<x<1$ and $0<y<1$.
Explanation of Method
The integrand $\frac{x^2}{1+y^2}=x^2\cdot\frac{1}{1+y^2}$ is a product of a function of $x$ alone and a function of $y$ alone, so over the rectangle $[0,1]\times[0,1]$ the double integral factors into a product of two single integrals.
- Recognize the separable form.
- Evaluate $\int_0^1 x^2\,dx$.
- Evaluate $\int_0^1 \frac{1}{1+y^2}\,dy$ using $\arctan$.
- Multiply the results.
Step 1 · Identify the region
$D=[0,1]\times[0,1]$, the unit square.
Step 2 · Set up the integral
$$ I = \int_0^1\int_0^1 \frac{x^2}{1+y^2}\,dx\,dy. $$
Step 3 · Recognize separability
$$ \frac{x^2}{1+y^2} = x^2\cdot\frac{1}{1+y^2}, $$
so
$$ I = \left(\int_0^1 x^2\,dx\right)\left(\int_0^1 \frac{1}{1+y^2}\,dy\right). $$
Step 4 · Evaluate the first integral
$$ \int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac13. $$
Step 5 · Evaluate the second integral
$$ \int_0^1 \frac{1}{1+y^2}\,dy = [\arctan y]_0^1 = \arctan(1)-\arctan(0) = \frac{\pi}{4}. $$
Step 6 · Multiply the results
$$ I = \frac13\cdot\frac{\pi}{4} = \frac{\pi}{12}. $$
Additional Notes
- Checking for separability before setting up an iterated integral can save significant work — here it avoids any coupling between the $x$- and $y$-integrations.
Result
The solution to the integral is: