1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface Integrals of first octant

  1. Nov 11, 2014 #1
    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. jcsd
  3. Nov 11, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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) =~?##
     
  4. Nov 13, 2014 #3
    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
  5. Nov 13, 2014 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##
     
  6. Nov 13, 2014 #5
    Oh oops. I think it should be r(x,y) = 3i + 2j + k - 6, right?

    Hmm I shall try again. :)
     
  7. Nov 14, 2014 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    Does it look right?
     
  8. Nov 16, 2014 #7
    I tried to do this as I learnt how to convert cartesian equation of a plane into its vector parametric form at A levels:

    e64p3r.jpg
    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? :)
     
  9. Nov 16, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that is correct. Have you figured out the xy domain?
     
  10. Nov 16, 2014 #9

    Zondrina

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Surface Integrals of first octant
Loading...