Calculating line integral on a vector field, help me find the mistake Alright, so, I have vector field: $F=[p(x,y), q(x,y)]=[y^3+e^{x^2}, x^3+{\tan}^2y]$. I need to calculate $\oint_Lpdx+qdy$, where $L: x^2+y^2+4y=0$. I transform it to $x^2 + (y+2)^2 = 4$, i.e. a circle with $r=2$ with origin at $(0,-2)$. The circle is "positively oriented", so I guess the integral should be going in counterclockwise direction.
The lecture from which this example assignment comes from introduces line integral and Green's theorem. Given presence of $e^{x^2}$, using the theorem is a must.
Thus: $$\oint_Lpdx+qdy = \iint_D (\frac{\partial{q}}{\partial{x}}-\frac{\partial{p}}{\partial{y}})dxdy = \iint_D(3x^2-3y^2)dxdy$$
I then try to change cooridnates to polar.
\begin{equation}
    \begin{cases}
      x &= R\cos{\varphi} \\
      y + 2 &= R\sin{\varphi} \\
    \end{cases}       
\end{equation}
After subsituting the circle equation, $R=2$. Then I define $D$ area:
\begin{equation}
  D =
    \begin{cases}
      r \in (0, 4) \\
      \varphi \in (\pi, 2\pi) \\
    \end{cases}       
\end{equation}
Back to double integral:
$$\int_\pi^{2\pi}\int_0^4(3(2\cos\varphi)^2-3(2\sin\varphi-2)^2)rdrd\varphi$$
Aaaaand, apparently after verifying with WolframAlpha, I have already failed here. The answer after solving the double integral should be $72\pi$. I get something like $-382-288\pi$ (might be inaccurate, writing from memory). Can you help me find the problem? I suspect I'm being dumb about the polar coordinate substitution, or $r$ has wrong range, or Jacobian is wrong
 A: After your transformation $x=x'$ and $y=y'-2$ since $dxdy=dx'dy'$ the integral becomes $\int\int_{D'}3(x^2-(y-2)^2)dxdy$ where $D': 0\leq r\leq 2$, $0\leq\theta\leq 2\pi$. Hence, the integral is
$$\int_0^{2\pi}\int_0^23(r^2\cos2\theta+4r\sin\theta-4)rdrd\theta=\int_0^{2\pi}\int_0^2(-12)rdrd\theta=(-12)(2\pi)(\frac{2^2}{2})=-48\pi$$
...
Another way: The circle in polar coordinates is $r=-4\sin\theta$ and so the domain is $D: 0\leq r\leq -4\sin\theta$, $\pi\leq\theta\leq 2\pi$. So, the integral is
$$\int_\pi^{2\pi}\int_0^{-4\sin\theta}(3r^2\cos2\theta)rdrd\theta=192\int_\pi^{2\pi}\sin^4\theta\cos2\theta d\theta=192(-\frac\pi 4)=-48\pi$$
where I took some help.
A: The limits for $\;r\;$ are $\;0\le r\le -4\sin\phi\;$ , so the integral's
$$\int_\pi^{2\pi}d\phi\int_0^{-4\sin\phi}3r^3(\cos^2\phi-\sin^2\phi)dr=\int_\pi^{2\pi}d\phi\int_0^{-4\sin\phi}3r^3(\cos^2\phi-\sin^2\phi)\,dr=$$
$$=3\int_\pi^{2\pi}\cos2\phi\,d\phi\,\,\frac14(-4\sin\phi)^4=192\int_\pi^{2\pi}\cos2\phi\,\sin^4\phi\,d\phi=-
48\pi$$
