# Flux integral through elliptic cylinder

Find the flux of F=(3x,2y,z) through the volume bound by the xy plane, the elliptic cylinder (x/3)^2+(y/2)^2=1, and the paraboloid x^2 + y^2 =z, and hence find the components of the flux through the 3 individual surfaces.

I parametrised the whole thing with x=3rsin and y=2rcos (J=6r)

and by divergence theorem found the total flux to be 117pi

When I try to calculate the flux through the elliptic cylinder by the regular method,

with parametrisation (3cos, 2sin, z)

and the corresponding cross product (2cos, 3sin, 0)

and boundary for z is z=9-5sin^2 by plugging in the 2 equations

I get 202.5pi, which is greater than the total 117pi.

I did the math, and I got that the total flux on the whole closed surface is $117\pi$ too, but the flux the "lateral surface" $\mathcal L=\big\{(x,y,z)\in\mathbb R^3:\ \big(\frac{x}3\big)^2+\big(\frac{y}2\big)^2=1,\ 0\leq z\leq x^2+y^2\big\}$, aka the elliptic cylinder, is $225\pi$ according to my calculations. I think I made a mistake, but that's a minor issue: a flux can be negative too, so your result is alright.
The flux through the lower surface $\mathcal B=\big\{(x,y,z)\in\mathbb R^3:\ \big(\frac{x}3\big)^2+\big(\frac{y}2\big)^2\leq 1,\ z=0\big\}$ is zero, as the inner product between the field and the normal vector of $\mathcal B$ is $\vec F(x,y,z)\cdot\hat{n}(x,y,z)=z$, but $z=0$ for $(x,y,z)\in\mathcal B$.
So it boils down to the parabolic surface: I calculated it and got $-\frac{171}2\pi$, which if the flux through $\mathcal L$ is really $\frac{405}2\pi$ as you said, and not $225\pi$, is the right answer, as clearly $\frac{405}2\pi-\frac{171}2\pi=117\pi$.