r=1/(8cosθ) for −π/2≤θ≤π/2 and r=1. Write an iterated integral in polar coordinates representing the area inside the curve r=1 and to the right of r=1/(8cosθ)

Answer 1

The graph of the area in question is to the area of a half circle in the 1st and 4th quadrants but the function

r = 1/8cos(t) takes a strip out of the half circle at x = .125 where t = 0 (r = .125)

The area of the half circle is .5*(1)2 = 1.5708

The area of the strip is area integral (dy dx) with limits x = 0, .125 and y = -SQRT(1 - x²) to + SQRT(1 - x²)

The strip area is .24935

So the area in question is 1.5708 - .24935 = 1.3215