Double integral over an area

Question

Find the double integral of $x$ over $S$ where $S$ is the area bounded by $x^2 + y^2 \leqslant 2$ and $x \geqslant 1$. Use the following subtitution: $x=r\cosθ ,y=r\sinθ$.

I have separated the original integral (let it be $I$) into $I=J-K-2L$, where $J$ is the double integral from $-π/2$ to $π/2$ and from $0$ to$\sqrt2$ of $r\cosθ$, $K$ is the double integral from $π/4$ to $π/2$ and from $0$ to $\sqrt2$ of $r\cosθ$, $L$ is the double integral from $0$ to $1$ and from $0$ to $x$ of $x$.

The result I got was: $(3\sqrt2 - 10)/6$ but I am not sure if it is correct because it is negative. Is my method correct? If not, which is?

Answer 1

Note that $x=1$ in polar coordinates is $r\cos \theta =1$ which is $ r=\sec \theta$

Thus the region of integration is simply $$ -\pi /4\le \theta \le \pi /4 \\ \sec \theta \le r \le \sqrt 2$$

You just need to integrate $r\cos \theta$ over that region.

Answer 2

Absolutely wrong. You have to calculate $$\iint_Sx\,\mathrm dx\mathrm dy,$$ where $x^2+y^2\le2$. Now this becomes$$2\iint_0^\sqrt{2-x^2}x\,\mathrm dx\mathrm dy=2\int x\sqrt {2-x^2}\,\mathrm dx.$$ You can calculate by taking $x$ from $0$ to $1$ but you need to substitute $\cos t$ and $\sin t$. Put $x=\sqrt 2 \cos t$ and $y=\sqrt 2 \sin t$ as $r=\sqrt 2$, then it becomes $$4\sqrt 2 \int_0^{2\pi}\cos^2t \sin t \,\mathrm dt.$$ Then apply gamma function to find $\dfrac{\sqrt 2}{3}$.