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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


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    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


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    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


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    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:

    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


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


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    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.
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