Double Integral with Polar Co-ordinates

Question

Could someone please help me work out this double integral with polar co-ordinates. I'm getting my answer to be 2cos(θ)/π. See the attached picture: double int. with polar co-ords

Answer 1

The answer can't depend on $\theta$.

Hint:

It is the integral of a function on the upper half of the disc delimited by the circle with center $(1,0)$ on the $x$-axis and radius $1$. The polar equation of this circle, which passes through the origin and has diameter $2$, is $$r=2\cos \theta,\quad \text{ by the inscribed angle theorem.}$$

Answer 2

$$y>0 \Rightarrow 0 < \theta < \pi$$ $$(x-1)^{2}+y^{2} < 1 \Rightarrow x^{2}+y^{2} < 2x \Rightarrow r^{2} < r \ cos\theta \Rightarrow r < cos \theta$$

Thus $$I = \int_{0}^{\pi}\int_{0}^{cos\theta}\frac{rcos\theta}{r^2}r \ dr \ d\theta = \int_{0}^{\pi}\int_{0}^{cos\theta}cos\theta \ dr \ d\theta = \int_{0}^{\pi} cos^{2}\theta= \frac{sin(2 \pi)}{4}+\pi=\pi$$