Exercise · ZW-POLAR-COORDINATES-009 · ★★★ Advanced
Integrating \(e^{-(x^2+y^2)}\) over the disc of radius \(a\) and letting \(a\to\infty\), the value of \(\iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dA\) is:
Choose one option.
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- B. \(\pi\)
Exercise · ZW-POLAR-COORDINATES-009 · ★★★ Advanced
Choose one option.
\(\int_0^{2\pi}\!\int_0^a e^{-r^2}r\,dr\,d\theta=\pi(1-e^{-a^2})\to\pi\). Since this is the square of \(\int_{-\infty}^{\infty}e^{-x^2}dx\), that integral equals \(\sqrt\pi\).
Polar coordinates locate a point of the plane by its distance \(r\) from the origin and the angle \(\theta\) its radius makes with the positive \(x\)-axis.