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From schaum's vector analysis: enter image description here

I project the differential area $dS$ of the plane onto the $xy$ plane, then

$dxdy = dS (|\hat n. \hat k|)$ Where $\hat n$ is the normal vector to $dS$

then $dS = \frac{dxdy}{ |\hat n. \hat k| }$

$ \hat n = \frac{ \nabla S}{ | \nabla S | } = \frac{2}{3} \hat i + \frac {1}{3} \hat j + \frac{2}{3} \hat k$

$\nabla\times\vec F = 3 \hat i - \hat j - 2 \hat k$

$(\nabla\times\vec F) . \hat n = \frac{1}{3} $

$dS = \frac{3}{2} dxdy$

Then, $\iint_S \nabla\times\vec F \cdot\ \hat n $ $dS$ = $ \frac{1}{2} \iint_S dxdy = \frac{1}{2} \iint_0^2 dxdy$ = $ \frac{1}{2} \int_0^1 2 dx = 1$

Now this is is when the boundaries are $x=0, x=1, y=0, y=2$

But when the boundaries are $x=0 , y=0, z =0$ , as follows:

enter image description here

It's all the same steps except for:

$\iint_S \nabla\times\vec F \cdot\ \hat n = \frac{1}{2} \iint_S dxdy = \frac{1}{2} \iint_0^{6-2x} dxdy = \frac {1}{2} \int_0^3 6-2x dx = \frac {9}{2}$

Now what I don't understand is: In the first part, we treated $y$ as changing independently of $x$ from $y=0$ to $y=2$

In the second part, we treated $y$ as dependent on $x$ by the function $y=6-2x$ and integrated from $y=0$ to $y=6-2x$

Why is that? When do we substitute $y$ in as a function of $x$ and integrate like in the second part, and when not to? The problem is I cannot visualize 3 planes intersecting each other and visualize $S$ , so how can I understand it?

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In the first problem, you are integrating the surface of the plane $\mathcal{P}: 2x+y+2z=6$ over the rectangle $0\leq x\leq 1, 0\leq y\leq 2$. To parametrize the rectangle, simply use $x,y$ as parameters and integrate as you did.

For the second problem, you are integrating the surface of $\mathcal{P}$ above the $xy-$plane. Imagine the plane $\mathcal{P}$ as a sheet of paper slicing into the $xy-$plane. The sheet creates a shadow over the $xy-$plane, which will be in the shape of a triangle. To see this, let $z=0$ to see what the plane looks like in the $xy-$plane. We'll get $$2x+y=6$$ Now, this is the line $y=6-2x$ which, when bounded by the lines $x=0$ and $y=0$, gives us a triangle. The triangle is a simple region and can be parametrized, for example, by $$0\leq x\leq 3$$ $$0\leq y\leq 6-2x$$ which explains the solution to the second problem.

To address your point about $y$ being "independent" from $x$ in the first integral and not the second, it's because the first region is a rectangle, which can be parametrized by all constant bounds (this is the easiest case). However, for more general regions (like the triangle in problem $2$), you cannot express them as $$a\leq x\leq b$$ $$c\leq y \leq d$$ but for example $$a\leq x\leq b$$ $$f(x)\leq y\leq g(x)$$ where $y$ is bounded by two continuous functions of $x$. In this problem, those two functions are $f(x)=0$ and $g(x)=6-2x$.

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  • $\begingroup$ Thank you, I got the idea and it's pretty intuitive now, however the plane when it slices the xy plane and casts a shadow on the xy plane is a bit hard to visualize, so eliminating z and working with $y=6-2x$ on the xy plane is easier and makes me see the triangle, but, can I do this in every problem? Eliminate z to see the shadow, right? [Provided that the boundaries are the $x=0$ , $y=0$ planes, ofcourse $\endgroup$
    – khaled014z
    Dec 25 '18 at 23:36
  • $\begingroup$ Well, given any surface $z=f(x,y)$, you can always set $z=0$ to see what the surface looks like in the $xy-$plane. This is sometimes called a "slice" or "cross-section" of the surface. The fact that your surface is a plane, combined with the fact that the other boundaries are $x=0$ and $y=0$, is why we get a triangle. As another example, consider the paraboloid $z=x^{2}+y^{2}-4$. When $z=0$, the corresponding curve in the $xy-$plane is the circle $x^{2}+y^{2}=4$. Looking at the graph of the surface, you could probably guess that this cross-section will be circular. $\endgroup$
    – pwerth
    Dec 25 '18 at 23:41
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Drawing boundaries on a graph helps a lot to understand the limits of the integrand. For second part Q67), after drawing the boundaries on a xy plane, we get intuition for the boundaries pretty well, https://www.desmos.com/calculator/19sfc9hohv ,here vertical strip changes from y=0 to y=-2x+6 and that strip stretches within x=0 to x=3.

But for first part Q66), after drawing the boundaries on xy plane, we get a rectangular region, where y(plane in 3D but a straight line in 2D) is obviously(intuition from graph) independent of x. Vertical strip changes from y=0 to y=2 and stretch that within x=0 to x=1 graph at https://www.desmos.com/calculator/xnho1f9uqu

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