Double integrals in polar coordinates
Problem 11.36 · medium
Evaluate \( \displaystyle \iint_D e^{- x^{2} - y^{2}} \, dA \) over the disk \( \displaystyle x^2 + y^2 \le 9 \) using polar coordinates.
- In polar coordinates the disk is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ.
- \[ e^{- r^{2} \sin^{2}{\left(\theta \right)} - r^{2} \cos^{2}{\left(\theta \right)}} = e^{- r^{2}} \]The integrand in r and θ.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{3} r e^{- r^{2}}\, dr\, d\theta = - \frac{\pi}{e^{9}} + \pi \]Integrate, remembering the extra r.✓ Proved
Answer \( - \frac{\pi}{e^{9}} + \pi \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the integral done in x and y, numerically, without polar coordinates, agrees |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/double_integral_polar, checked 2026-09-26 with SymPy 1.14.0.