I need help calculating $I$ where:
$$I = \iint_R\frac{dxdy}{\sqrt{1-x^2-y^2}}
,\qquad R = \left\{ (x,y) \in \Bbb R^2 : x^2 + y^2 -x \le 0 , y \ge 0 \right\}$$
The possible answers are
- $\:\pi$
- $\:2\pi-1$
- $\:2\pi+2$
- $\:\pi +1$
I have converted it to polar where I ended up with the integral:
$$I = \int_0^\pi\int_0^{cos{\theta}}\frac{r}{\sqrt{1-r^2}}\:dr\:d\theta$$
I have tried evauluating it but the answer I get is: $\:\pi -2$
I took the following steps:
- $\int_0^{\cos\theta}\frac{r}{\sqrt{1-r^2}} ,\:u = 1-r^2,\: dr = \frac{-1}{2r}du$
- $\int_0^{\cos\theta}\frac{r}{\sqrt{1-r^2}} =\int \frac{r}{\sqrt{u}}\frac{-1}{2r}du = \frac{-1}{2}\int\frac{1}{\sqrt{u}}du = \frac{-1}{2}\left[2\sqrt{1-r^2} \right]_0^{\cos\theta} = -\sqrt{1-\cos^2\theta} \:\:+ 1 $
- $-\sqrt{1-\cos^2\theta} \:\:+ 1 = 1 – \sin\theta$
- $\int_0^\pi 1- \sin\theta \:d\theta = \int_0^\pi1\:d\theta + \int_0^\pi -\sin\theta \:d\theta$
- $\int_0^\pi1\:d\theta = \pi$
- $\int_0^\pi -\sin\theta \:d\theta = \left[ \cos\theta\right]_0^\pi = \cos(\pi) – \cos(0) = -2$
- From steps 4, 5 and 6 we get $\:I = \pi -2$
Somewhere along these steps I have made a mistake and I can't find it.
Best Answer
HINT
To begin with, notice that: \begin{align*} R & = \left\{(x,y)\in\mathbb{R}^{2} \mid \left(x - \frac{1}{2}\right)^{2} + y^{2} \leq \frac{1}{4}\right\}\cap\{(x,y)\in\mathbb{R}^{2} \mid y\geq 0\} \end{align*} which can be written in polar coordinates as (according to the cosine's law): \begin{align*} R = \left\{(r,\theta)\in\mathbb{R}_{\geq 0}\times[0,2\pi) \mid (0\leq r\leq \cos(\theta)) \wedge (0\leq \theta\leq \pi/2)\right\} \end{align*}
Consequently the proposed integral corresponds to: \begin{align*} \iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\sqrt{1 - x^{2} - y^{2}}} & = \int_{0}^{\pi/2}\int_{0}^{\cos(\theta)}\frac{r}{\sqrt{1 - r^{2}}}\mathrm{d}r\mathrm{d}\theta\\\\ \end{align*} whence I guess you can proceed.