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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.

Ans:

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

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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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