# Surface Integrals of first octant

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1. Nov 11, 2014

### galaxy_twirl

1. The problem statement, all variables and given/known data

Evaluate ∫∫ F⋅dS, where F = yi+x2j+z2k and S is the portion of the plane 3x+2y+z = 6
in the first octant.

The orientation of S is given by the upward normal vector.

2. Relevant equations

∫∫S F⋅dS = ∫∫D F(r(u,v))⋅||ru x rv|| dA, dA=dudv

3. The attempt at a solution

Since this is the first octant, our domain will be 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

I have to obtain the equation of the form r(u,v) before I proceed to substitute it into the equation given by F. However, I am stuck trying to obtain the equation r(u,v).

Just wondering, is r(u,v) here the vector equation of the plane? (Cos r(u,v) is usually in the form i, j, k). From deduction, I found that x=y=z=1.

Thank you for any help given. :)

2. Nov 11, 2014

### LCKurtz

No it won't. This problem has nothing to do with circles or $\pi$. It is a plane slanted up against the first octant coordinate planes and its xy domain is not a rectangle. Have you drawn a sketch?

Let $u=x,~v=y$ be your parameters. So then $\vec r(x,y) =~?$

3. Nov 13, 2014

### galaxy_twirl

Oh nope, I haven't drawn a sketch. I think I misread the question. How do I sketch the intersection? I know how to sketch the octant, which is like a quarter of a 3D sphere, but I am unsure of how to draw the plane. May I have your advice on sketching the plane? :)

r(x,y) = 3u i + 2v j + k = 6, am I right to say this?

Thanks! :)

Last edited by a moderator: Nov 14, 2014
4. Nov 13, 2014

### LCKurtz

Nothing in this problem is "like a 3D sphere". Find the x,y, and z intercepts. Join them with straight lines in the coordinate planes, and that will show you the first octant portion of the plane.

No. Not even close. You have x and y on the left and u and v on the right. Like I said before, your parameters are x and y. Your parameterization should look like $r(x,y) = xi + yj + ? k$

5. Nov 13, 2014

### galaxy_twirl

Oh oops. I think it should be r(x,y) = 3i + 2j + k - 6, right?

Hmm I shall try again. :)

6. Nov 14, 2014

### LCKurtz

Does it look right?

7. Nov 16, 2014

### galaxy_twirl

I tried to do this as I learnt how to convert cartesian equation of a plane into its vector parametric form at A levels:

May I know if what I had done is correct?

Assuming it is correct, I guess my parametization of the plane should be:

r(x,y) = xi + yj + (6-3x-2y)k

right? :)

8. Nov 16, 2014

### LCKurtz

Yes, that is correct. Have you figured out the xy domain?

9. Nov 16, 2014

### Zondrina

Your parametrization looks fine. So $\vec r(x,y) = x \hat i + y \hat j + z \hat k$ where $z = 6 - 3x - 2y$.

Looking at the theorem you provided in the given equations:

You should notice you are using the wrong equation. In-fact, your normal vector has completely disappeared unless you intended to mean:

$$\iint_S \vec F(x, y, z) \cdot d \vec S = \iint_S \vec F(x, y, z) \cdot \vec n \space dS = \iint_D [\vec F(\vec r(x,y)) \cdot \vec n] \space ||\vec r_x \times \vec r_y|| \space dA$$

It can be shown with $\vec n = \frac{\vec r_x \times \vec r_y}{||\vec r_x \times \vec r_y||}$ that the surface integral can be evaluated in the following fasion:

$$\iint_S \vec F(x, y, z) \cdot d \vec S = \iint_D \vec F(\vec r(x,y)) \cdot (\vec r_x \times \vec r_y) \space dA$$

which is much simpler in many cases.