극좌표 이중적분

(a) On the domain of integration

• $y$ runs from $0$ to $1$. In inequalities, $0\le y\le 1$.
• For each fixed $y$ in that range, $x$ runs from $\sqrt{3}\,y$ to $\sqrt{4-y^2}$. In inequalities, that is $\sqrt{3}\,y\le x\le \sqrt{4-y^2}$. Note that the inequalities $x\le \sqrt{4-y^2}$, $x\ge 0$ are equivalent to $x^2+y^2\le 4$, $x\ge 0$.

Note that the line $x=\sqrt{3}\, y$ and the circle $x^2+y^2\le 4$ intersect when $3y^2+y^2=4$, i.e. $y=\pm 1$. Here is a sketch.

(b) In polar coordinates, the circle $x^2+y^2= 4$ is $r=2$ and the line $x=\sqrt{3}\, y$, i.e. $\frac{y}{x} =\frac{1}{\sqrt{3}}$, is $\tan\theta=\frac{1}{\sqrt{3}}$ or $\theta=\frac{\pi}{6}$. As $\mathrm{d}x\,\mathrm{d}y = r\,\mathrm{d}r\,\mathrm{d}\theta$, the domain of integration is

$$\begin{aligned} \big\{ \ (r\cos\theta,r\sin\theta)\ \big|\ 0\le \theta\le{\pi}/{6},\ 0\le r\le 2\ \big\} \end{aligned}$$

and

$$\begin{aligned} \int_0^1 \int_{\sqrt{3}y}^{\sqrt{4-y^2}} \ln\big(1+x^2+y^2)\ \mathrm{d}x\,\mathrm{d}y &= \int_0^2\mathrm{d}r\int_0^{\pi/6}\mathrm{d}\theta\ r\,\ln(1+r^2) = \frac{\pi}{6}\int_0^2\mathrm{d}r\ r\,\ln(1+r^2) \\ &=\frac{\pi}{12}\int_1^5 \mathrm{d}u \ln(u) \qquad\text{with }u=1+r^2,\ \mathrm{d}u=2r\,\mathrm{d}r \\ &=\frac{\pi}{12}\Big[u\ln(u)-u\Big]_1^5 \\ &=\frac{\pi}{12}\big[5\ln(5)-4\big] \end{aligned}$$