# bounds for surface integral of a plane

I need to calculate the next surface integral, but i'm having troubles with the bounds;

$$\iint -2 \, dS,$$

where $S$ is the part of the plane $x+2y+z=2$ that is cut off in the first octant.

My function is $\vec{\rm F}\left(x,y,z\right)=\left(y - x,x - z,x - y\right)$. I found the $-2$ with Stokes' theorem and i'm sure that part is correct. Now I still have to prove that the surface integral is $1$. Can someone please help me ?.

• So what you want to actually compute is $\int_C F \cdot d \vec{r}$, using Stokes's theorem? – Mark Fantini Dec 21 '13 at 22:10
• I.just want to calculate the surface integral for the constant 1 function to prove that it's 1 (i already know that that's the correct answer) but i don't know how to pick the bounds – user2162627 Dec 21 '13 at 22:12
• What you have is a vector field, not a real-valued function. Either way, to find the bounds you can isolate $z$ in terms of $x,y$, use the formula for the normal vector of a function, and the region in the plane is found by setting $z=0$ in the plane equation. However, I strongly suggest you write the problem as stated. This is confusing. – Mark Fantini Dec 21 '13 at 22:16
• What do you mean 'write the problem as stated'? My English is not so good, i don't understand. – user2162627 Dec 21 '13 at 22:18
• Is the exercise in English? What I meant is to copy the problem and paste it here, as it is written. – Mark Fantini Dec 21 '13 at 22:21

The curl of your vector field is

$$\nabla \times \vec{F} = (0,-1,0).$$

Therefore, by Stokes's theorem we know that

$$\int\limits_{C} \vec{F} \cdot d \vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot d \vec{S}.$$

Since this is a plane the normal is $(1,2,1)$. Plugging these yields

$$\iint\limits_{S} (\nabla \times \vec{F}) \cdot d \vec{S} = \iint\limits_{D} (-2) \, dA = -2 \iint\limits_{D} \, dA.$$

I'm denoting by the $D$ the area in the plane. This means that circulation of $\vec{F}$ is $-2$ times the area in the plane projected by the surface.

Whenever we integrate over surfaces the basic idea is to parametrize it by a region in $\mathbb{R}^2$. When we perform the integration, the idea is to "pull back" the integration on the surface to integration in $\mathbb{R}^2$, which is what we know.

This is a slanted plane, but if you managed to view it from the $z$ axis you would not be able to distinguish it from a triangle in the $xy$ plane. That is the projected area. It is the region in the plane bound by

$$\begin{cases} x \geq 0, \\ y \geq 0, \\ x+2y \leq 2. \end{cases}$$

The last inequality was found setting $z=0$ in the plane equation, giving the boundary. The area of this triangle is simple: it has $2$ units as base and $1$ as height, therefore

$$\int\limits_{C} \vec{F} \cdot d \vec{r} = -2 \iint\limits_{D} \, dA = -2 \cdot \left( \frac{1}{2} \cdot 2 \cdot 1 \right) = -1.$$

Minus sign is due to orientation. I believe we used the counterclockwise orientation all along.

• Thank you so much again! You're so kind to explain it this well. You've really helped me understand the principle! – user2162627 Dec 21 '13 at 23:25
• You are welcome. I'm glad to have helped. :) – Mark Fantini Dec 21 '13 at 23:26