Evaluate the flux of the vector field without using divergence theorem

Given $F(x,y,z)=(z^2-x,-xy,3z)$ and $S$ is the surface of the solid delimited by the equations $z=4-y^2, x=0, x=3$ and $z=0$, with the normal vector exterior, evaluate $\iint_S F\cdot n \,ds$ without using The Divergence Theorem.

• Not that it mattered, but $n$ on the bottom should be $(0,0,-1)$. The first integral should be $48$. That should fix it. – Ted Shifrin Aug 26 '17 at 18:46
• @TedShifrin about the bottom, i'll edit it. About the first integral, why? I'm not finding the error. – Gustavo Alves de Oliveira Aug 26 '17 at 18:50
• It appears you dropped an exponent on the $-3y^2$. :P – Ted Shifrin Aug 26 '17 at 18:54
• Yes! I just saw this. Thanks btw. – Gustavo Alves de Oliveira Aug 26 '17 at 18:54 Here's what I tried:

I firstly diveded the surface in 4 pieces: a,b,c and d. Then I worked with then separetely. The result is the sum of all integrals.

Cylinder surface (a):

$\alpha(x,y)=(x,y,4-y^2)$, with $0<x<3$ and $-2<y<2$

$\alpha_x = (1,0,0)$ and $\alpha_y = (0,1,-2y)$

So, $N=(0,2y,1)$ and $F(\alpha(x,y))=((4-y^2)^2-x,-xy,3(4-y^2))$.

Then $\iint_a F \cdot n \,ds=\iint F(\alpha(x,y)) \cdot N \,dx\,dy=\iint-2xy^2+12-3y^2 \,dx\,dy = 48$

Buttom surface (b):

$\alpha(x,y)=(x,y,0)$ with $0<x<3$ and $-2<y<2$

$F(\alpha(x,y))=(-x,-xy,0)$ and $N=(0,0,-1)$

So, $\iint_b F \cdot n \,ds =0$

Cover with x=0 (c):

$\alpha(y,z)=(0,y,z)$ with $-2<y<2$ and $0<z<4-y^2$

$F(\alpha(y,z))=(z^2,0,3z)$ and $N=(-1,0,0)$

So, $\iint_c F \cdot n \,ds = \iint F(\alpha(y,z)) \cdot N \,dy\,dz = \iint -z^2 \,dy\,dz=-4096/105$

Cover with x=3 (d):

$\alpha(y,z)=(3,y,z)$ with $-2<y<2$ and $0<z<4-y^2$

$F(\alpha(y,z))=(z^2-3,-3y,3z)$ and $N=(1,0,0)$

So, $\iint_c F \cdot n \,ds = \iint F(\alpha(y,z)) \cdot N \,dy\,dz = \iint z^2-3 \,dy\,dz=736/105$

Thus,

$\iint_S F \cdot n \,ds = 48+0+736/105-4096/105=16$