1. **Stating the problem:** We need to evaluate the double integral $$\int_0^{\frac{\pi}{2}} \int_0^{\sec \theta} r^2 \cos \theta \, dr \, d\theta$$. 2. **Understanding the integral:** The integral is in polar coordinates where $r$ ranges from $0$ to $\sec \theta$ and $\theta$ ranges from $0$ to $\frac{\pi}{2}$. The integrand is $r^2 \cos \theta$. 3. **Integrate with respect to $r$ first:** $$\int_0^{\sec \theta} r^2 \cos \theta \, dr = \cos \theta \int_0^{\sec \theta} r^2 \, dr = \cos \theta \left[ \frac{r^3}{3} \right]_0^{\sec \theta} = \cos \theta \cdot \frac{(\sec \theta)^3}{3} = \frac{\cos \theta}{3} \sec^3 \theta.$$ 4. **Simplify the expression:** Recall $\sec \theta = \frac{1}{\cos \theta}$, so $$\frac{\cos \theta}{3} \sec^3 \theta = \frac{\cos \theta}{3} \cdot \frac{1}{\cos^3 \theta} = \frac{1}{3 \cos^2 \theta}.$$ 5. **Now integrate with respect to $\theta$:** $$\int_0^{\frac{\pi}{2}} \frac{1}{3 \cos^2 \theta} \, d\theta = \frac{1}{3} \int_0^{\frac{\pi}{2}} \sec^2 \theta \, d\theta.$$ 6. **Recall the integral formula:** $$\int \sec^2 \theta \, d\theta = \tan \theta + C.$$ 7. **Evaluate the definite integral:** $$\frac{1}{3} \left[ \tan \theta \right]_0^{\frac{\pi}{2}} = \frac{1}{3} (\tan \frac{\pi}{2} - \tan 0).$$ 8. **Evaluate the limits:** $\tan 0 = 0$, but $\tan \frac{\pi}{2}$ tends to infinity, so the integral diverges. **Final answer:** The integral diverges to infinity because $\tan \frac{\pi}{2}$ is not finite.