Double Integrals in Polar Coordinates — Question 5

Question 5 – Solution

Strategy. The exponent and disk are radial; use $u=r^2$ after including the Jacobian.

Step 1: Set up $$I(a)=\int_0^{2\pi}\int_0^a e^{-r^2}r\,dr\,d\theta.$$

Step 2: Evaluate With $u=r^2$, $du=2r\,dr$, $$I(a)=2\pi\left[-\frac 12e^{-r^2}\right]_0^a =\boxed{\pi(1-e^{-a^2})}.$$

Step 3: Checks Since $I'(a)=2\pi a e^{-a^2}>0$, enlarging the disk increases the integral. Also $$\boxed{\lim_{a\to\infty}I(a)=\pi}.$$ The result approaches a finite limit because the exponential decay dominates the growing circumference.