Calculating area defined by curves $L_1: y=-x \text{, } L_2:y^2+x^2=4 \text{ and } L_3:x=0$ using multible integrals and Polar coordinates

Question

I need to find an area that is formed by three curves: $L_1: y=-x \text{, } L_2:y^2+x^2=4 \text{ and } L_3:x=0$. I understand that these curves make two different areas, so I am trying to compute both of them.

So, as I understand the integrals that need to be computed is $$\int^{\frac{3\pi}{4}}_{\frac{-\pi}{4}}\int^2_0r^3sin\theta cos\theta drd\theta=4\int^{\frac{3\pi}{4}}_{\frac{-\pi}{4}} sin\theta cos\theta d\theta =2(sin^2(\theta)\left.\right|^{\frac{3\pi}{4}}_{-\frac{\pi}{4}}) =2(\frac{\sqrt{2}}{{2}}+\frac{\sqrt{2}}{{2}})=2\sqrt{2}$$

Is it correct or have I made a mistake?

The second area:

$$\int^{\frac{\pi}{2}}_{\frac{3\pi}{4}}\int^2_0 r^3sin\theta cos\theta = 2(sin^2(\theta)\left.\right|^{\frac{\pi}{2}}_{\frac{3\pi}{4}})=2(1-\frac{\sqrt{2}}{2})=2-\sqrt{2}$$

Answer 1

There are four finite regions.

The region with (polar) angles between $-\pi/4$ and $\pi/2$ has area $$ \int_0^2 \int_{-\pi/4}^{\pi/2} r \,\mathrm{d}\theta \,\mathrm{d}r = \int_0^2 \frac{3\pi}{4} r \,\mathrm{d}r = \left. \frac{3\pi}{4} \frac{r^2}{2} \right|^2_0 = \frac{3\pi}{2} \text{.} $$

The areas of the other three regions are entirely analogous.

Answer 2

I'm assuming the region is defined as:

Why work so hard with integrals? The area is the sum of a quarter of a circular disk of radius $2$ and a right triangle with sides equal $2$:

$$A = \frac{1}{4} \pi 2^2 + \frac{1}{2} 2 \cdot 2$$

The first portion is trivial in polar coordinates:

$$A_1 =\int\limits_{\theta = 0}^{\pi/4} 2 \theta d\theta$$

The second is only slightly more tricky:

$$\int\limits_{\theta = 0}^{-\pi/4} \frac{2 \theta}{\cos \theta}\ d\theta$$