Integral calculated directly and with Gauss Green formula

Question

I checked on Wolfram Alpha that the directly calculated integral was correct, when I go to use the Gauss Green formula, I get $0$ as a result, when instead, I should get $\left(4\pi-5\sqrt{3}\right)/3$.

• Could someone help me understand where I'm going wrong? I would really appreciate it.
• Calculate the following double integral ( directly and with the Gauss Green formula ). $$ \iint_D\left(x^2+y^2\right)dxdy, \quad D = \left\{\, 1\leq x^2+y^2 \leq 4,\ y \geq \sqrt{3}\,\right\} $$ $$ \mbox{with}\quad \left\{\begin{array}{rcl} x & = &r\cos(\theta) \\ y & = & r\sin(\theta) \\ dx\,dy & = & r \ dr \ d\theta \\ x^2+y^2 & = & r^2 \end{array}\right. $$

I have $P(r,\theta) = (2, \pi/3)$, and $O(r,\theta) = (2,2\pi/3)$. The arc $OQP$ corresponds to $r = 2$, the segment $OP$ is determined by setting $r \sin \theta = y = \sqrt{3}$ and then solving for $r$ as a function of $\theta$:$$r = \frac{\sqrt{3}}{\sin\theta}.$$Directly:$$\int_{\theta = \pi/3}^{2\pi/3} \int_{r=\frac{\sqrt{3}}{\sin\theta}}^2 r^3 \, dr \, d\theta = \frac{4\pi-5\sqrt{3}}{3}$$ With the Gauss Green formula: $$\iint\limits_D {\left( {{{\partial Q} \over {\partial x}} - {{\partial P} \over {\partial y}}} \right)da} = \int\limits_{\partial D} {Pdx + Qdy}$$ I choose $P = 0$
$$\iint_D \frac{\partial Q}{\partial x} \mathrm{da} = \int_{\partial D} Q \ \mathrm{dy}$$ so now I have to find a $Q$ such that $$\frac{\partial Q}{\partial x} = x^2+y^2$$ so
$$Q=\int x^2+y^2 \ \mathrm{dx} = \frac{x^3}{3}+xy^2$$ Now by the green theorem:
$$\iint_D x^2+y^2 \ \mathrm{da} = \int_{\partial D} \frac{x^3}{3}+xy^2$$

$$O(-1,\sqrt{3}), \ P(1,\sqrt{3}), \ Q(0,2)$$

The integral breaks into 3 parts :

$$\underbrace{\int_{OP} \frac{x^3}{3}+xy^2}_0+\underbrace{\int_{PQ} \frac{x^3}{3}+xy^2}_{I_1}+\underbrace{\int_{QO} \frac{x^3}{3}+xy^2}_{I_2}$$

$x^2+y^2 = 4 \Rightarrow x = \sqrt{4-y^2}$

$$I_1=\int_{y=\sqrt{3}}^2 \frac{ \sqrt{4-y^2}^3}{3}+ \sqrt{4-y^2}y^2dy = \frac{1}{6}\left(4\pi-5\sqrt{3}\right)\tag{1}$$

$$I_2=\int_{y=2}^{\sqrt{3}} \frac{ \sqrt{4-y^2}^3}{3}+ \sqrt{4-y^2}y^2dy = \frac{1}{6}\left(5\sqrt{3}-4\pi\right)\tag{2}$$ Sum up, i get $0$ and not the result $\frac{4\pi}{3}-\frac{5\sqrt{3}}{3}$.$$\frac{1}{6}\left(4\pi-5\sqrt{3}\right) + \frac{1}{6}\left(5\sqrt{3}-4\pi\right) = 0$$I noticed that if it was $-$ instead of $+$, I got my result exactly:$$\frac{1}{6}\left(4\pi-5\sqrt{3}\right) - \frac{1}{6}\left(5\sqrt{3}-4\pi\right) = \frac{4\pi-5\sqrt{3}}{3}$$.

Answer 1

Divide the integral in two pieces, first integrate along the segment $OP$ and then along the arc $PO$. The first integral is $0$ as you said. Now, parametrize the arc $PO$ as follows: $$\gamma(t)=(-t,\sqrt{4-t^2})\quad \gamma'(t)=\left(-1,-\frac{t}{\sqrt{4-t^2}}\right)\quad \text{for } t\in [-1,1] \quad $$ Hence your integral becomes: $$\int_{-1}^1 -\left(-\frac{t^3}3-t(4-t^2)\right)\frac{t}{\sqrt{4-t^2}}\mathrm dt=\frac{4\pi}3-\frac5{\sqrt 3} $$ according to Wolfram.