Evaluate the surface integrals using divergence theorem

$$ {\oint \oint}_S (xy\bar{i} + z^2 \bar{k}) \bar{n} dS $$

where S is the surface enclosing the volume in the first octant bounded by the planes $z= 0, y = x, y = 2x, x + y+ z = 6$ and $\bar{n}$ is the unit outer normal to S.


So by divergence theorem:

$${\oint \oint}_S (xy\bar{i} + z^2 \bar{k}) \bar{n} dS = \int \int_{\Omega} (y + 2z) dV $$

now adding the respective integral limits

$$\int_{0}^{2} \int_{x}^{2x} \int_{0}^{6 - x - y} (y - 2z) dz dy dx + \int_{2}^{3} \int_{x}^{6-x} \int_{0}^{6 - x - y} (y - 2z) dz dy dx$$

so the domain for each variable is

$0 \leq z \leq 6 -x - y$ they just rearrange the equation with respect to $z$

$x \leq y \leq 2x$ (this was given in the question)

$0 \leq x \leq 2$ (I don't understand how they got this could someone explain. Thank you.)

for the second integral:

$0 \leq z \leq 6 -x - y$ they just rearrange the equation with respect to $z$

$x \leq y \leq 6-x$ since $z$ is done with we can consider $z = 0$ and solve for $y$.

$2 \leq x \leq 3$ (I don't understand this either)

Not trying to solve the integral just wondering how they got the domain


The volume of integration is delimited by vertical planes and only two non vertical ones, this makes things very clear for the $z$ part of the integral (as indeed you checked). So, we need to understand the limits for $x$ and $y$ into the $z=0$ plane. These are determined by the intersection of three planes with the plane $z=0$, so is, these three lines

$y=x$, $y=2x$ and $x+y=6$. They define a triangle with vertex at $(0,0)$, $(3,3)$ and $(2,4)$

Now, we have to describe de range for the variation of $y$ depending on the value of $x$ we are considering.

If $0\leq x\leq 2$ then $x\leq y\leq 2x$

If $2< x\leq 3$ then $x\leq y\leq 6-x$

A picture helps to visualize.

Plane z=0


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