I have a doubt regarding evaluation of the following triple integral:
$$I=\iiint_V (4xy-2y+8xz) \, dV,$$ where $V$ is first octant of $y^2+z^2=9$ and planes $x=0, x=2$.
Given that the cylinder is symmetrical about $y =0, z=0$ planes and given that $I =\frac{1}{4} $ {volume over given cylinder and given planes } over $(4xy-2y+8xz)$.
As the given cylinder is symmetrical about $y= 0$ and $z=0$, if we put $-y$, $-z$ in $(4xy-2y+8xz)$ we get $(-4xy+2y-8xz)$ which means due to symmetry can I say it is zero.
But if I calculate using cylindrical coordinates I am getting $180$
Can you pls point me to where I am making error?
Think of what the words "first octant" means. The integration limits are not symmetric. Both $y$ and $z$ limits are from $0$ to $3$