# Surface integral confusion about boundaries

From schaum's vector analysis: 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: 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?

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 $$g(x)=6-2x$$.
• 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 Dec 25 '18 at 23:36
• 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. Dec 25 '18 at 23:41