Question

Sketch the region of integration, carefully with appropriate labels, and then perform the following integral: (0 to pi/4) ∫ (4secƟ to 0) ∫ r dr dƟ

Answer 1

$$\int_0^{\pi/4}\int_{4\sec(\theta)}^0r\,dr\,d\theta=\int_0^{\pi/4}\bigg[r^2/2\bigg]_{4\sec(\theta)}^0d\theta\\ =\int_0^{\pi/4}-8\sec^2(\theta)d\theta\\ =-8[\tan(\theta)]_0^{\pi/4}=-8$$