Flux of $v=(3y_1-2y_2^2+y_3^2,3y_1^2-2y_2+y_3^2,3y_1^2-2y_2^2+y_3)$ through teardrop shape 
$v=(3y_1-2y_2^2+y_3^2,3y_1^2-2y_2+y_3^2,3y_1^2-2y_2^2+y_3)$
Let $\varphi:K\to R^3$, where $K=[0,2\pi]\times[0,1]$ and $\varphi$ is given by $\varphi(x) = \{ (\cos x_1)(x_2-x_2^3), (\sin x_1)(x_2-x_2^3), x_2^2+ \frac13 x_2^6 \}$. Let $\tilde Y =\varphi(K)$. Calculate flux of $v$ throught $\tilde Y$.

So phi is this funny shape right here from what I understand:

I tried using the determinant method but I end up with these huge equations that I don't think I'll be able to solve. I also tried calculating the pullback, but again it's super long. Last idea I had was to maybe do a change of variables and set $u:=x_2-x_2^3$, but it doesn't really solve the problem.
How would you approach this? Detailed calculations of integrals in the explanation would be appreciated.
 A: Since the surface is closed, you may use the divergence theorem. In particular, if $V$ is the region bounded by $Y=\phi(K)$, then the flux of $v$ through $Y$ is given by
$$
\text{flux}= \int_Y v\cdot \text{ d} S= \int_{V}\text{div}(v)\text{ d}V= 2|V|,
$$
where $|V|$ denotes the volume of $V$.
We used the fact that the divergence of $v$ is
$$
\text{div}(v)= \frac{\partial}{\partial y_1}(3y_1-2y_2^2+ y_3^2)+ \frac{\partial}{\partial y_2}(3y_1^2-2y_2+y_3^2)+\frac{\partial}{\partial y_3}(3y_1^2-2y_2^2+y_3)=
2.
$$So we only need to compute the volume of the region enclosed by $Y$. Again, by the divergence theorem, 
$$
|V|= \int_V\text{ d}V= \int_{Y}(0, 0,y_3)\cdot\text{ d}S= \int_{K}(0, 0, x_2^2+\frac13 x_2^6)\cdot (\frac{\partial\phi}{\partial x_1}\times \frac{\partial\phi}{\partial x_2})\text{ d}x_1\text{ d}x_2,
$$
because $\text{div}(0,0,y_3)=1$. We compute,
$$
\frac{\partial\phi}{\partial x_1}\times \frac{\partial\phi}{\partial x_2}= \det\begin{pmatrix} i & j & k \\ -\sin(x_1)(x_2-x_2^3) & \cos(x_1)(x_2- x_2^3) & 0 \\ (1-3x_2^2)\cos(x_1) & (1-3x_2^2)\sin(x_1) & 2x_2+ 2x_2^5\end{pmatrix}= (\cos(x_1) 2x_2^2(1-x_2^2)(1+x_2^4), \sin(x_1)2x_2^2(1-x_2^2)(1+x_2^4), -x_2(1-x_2^2)(1-3x_2^2)),
$$
thus,
$$
|V|= \int_K -x_2(1-x_2^2)(1-3x_2^2)(x_2^2+\frac13 x_2^6)\text{ d}x_1\text{d}x_2 \= \int_{x_1=0}^{2\pi}\int_{x_2=0}^1 -x_2^3+ 4x_2^5-\frac{4}{3}x_2^7+\frac43x_2^9- x_2^{11} \text{ d}x_1\text{d}x_2= \frac{3\pi}{5}.
$$
As a result, given that my calculations are correct, the flux of $v$ through the surface $Y$ is $\frac{6\pi}{5}$.
