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Evaluating a surface integral using parametric/explicit representation

  1. Feb 4, 2013 #1

    xio

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    [EDIT]: Correct answer for this problem is 1/2, not 4 as I thought before; that means the result for the explicit representation was correct. Still I don't understand how to treat the case with the parametric representation.

    Greetings,

    I need to evaluate $$\iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS,$$ where ##S## is the plane surface whose boundary is the triangle with vertices at ##(1,0,0),(0,1,0),(0,0,1)## , ##\mathbf{F}(x,y,z)=(x,y,z)## , ##\mathbf{n}## is the unit normal to ##S## .

    There are two representations for the surface:

    • parametric ##\mathbf{r}(u,v)=(u+v,u-v,1-2u) ##
    • explicit ##z=f(x,y) ##

    and the general formula for this type of problems is $$\iint_{\mathbf{r}(T)}\mathbf{F}\cdot\mathbf{n}\, dS=\iint_{T}\mathbf{F}[\mathbf{r}(u,v)]\cdot\mathbf{n}\,\left\Vert \frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\Vert \, du\, dv ,$$ but since $$\mathbf{n}=\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}/\left\Vert \frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\Vert ,$$ it simplifies to $$\iint_{T}\mathbf{F}[\mathbf{r}(u,v)]\cdot\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\, du\, dv . $$

    Parametric representation:

    Given ##\mathbf{r}(u,v)=(u+v,u-v,1-2u) ##, we find $$\iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS=\iint_{T}(u+v,u-v,1-2u)\cdot(-2,-2,-2)\, du\, dv=-2\iint_{T}\, du\, dv. $$ The problem now is that I simply cannot figure out the region of integration ##T## : ##0\leq1-2u\leq1## suggests that ##u\in[0,0.5]## ; taking this into account and the fact that ##0\leq u+v\leq1## , ##v\in[0,0.5]## . But now the two conditions imply that ##-0.5\leq u-v\leq0.5## , while it should be ##0\leq u-v\leq1## .

    Explicit representation:

    My best guess for the explicit representation of the surface is ##z=-x-y+1## (I omit its derivation here for brevity).

    triangle.png

    Hence $$\iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS=\iint_{T}(x,y,-x-y+1)\cdot(1,1,1)\, dx\, dy=\iint_{T}\, dx\, dy .$$ Again I have troublems finding the region of integration. If it is the triangle on the ##z=0## plane, then ##x\in[0,1],y=1-x## and we have $$\iint_{T}\, dx\, dy=\int_{0}^{1}\left[\int_{0}^{1-x}dy\right]dx=1/2 ,$$ but the correct answer is “4” (I know it in advance).
     
    Last edited: Feb 4, 2013
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  3. Feb 4, 2013 #2

    vela

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    From the first two coordinates in the parametric representation, you have
    \begin{align*}
    x &= u+v \\
    y &= u-v
    \end{align*} Try solving for u and v in terms of x and y. You should be able to show that lines of constant u and constant v correspond to lines that make 45 degree angles with the x and y axes. In other words, this transformation is a 45-degree rotation (and some scaling).

    The projection of the surface onto the xy-plane is a triangle. Figure out what triangle this corresponds to on the uv-plane. It should be pretty straightforward to figure out the limits from that once you have that picture down.
     
  4. Feb 4, 2013 #3

    LCKurtz

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    The problem in finding the ##u,v## limits is to figure out what region in the ##uv## plane corresponds to the triangular domain in the ##xy## plane under your surface. That triangle is bounded by the lines ##x=0,\, y=0,\, x+y=1## in the ##xy## plane. Solving your two equations ##x=u+y,\, y = u-v## for ##x## and ##y## gives$$
    u = \frac{x+y} 2,\, v = \frac {y-x} 2$$Use these two equations to find what ##uv## region maps to your ##xy## region. You can use that ##uv## region to get your limits right. Try mapping the corners first. You will need to make some decision on the orientation to get the correct sign.

    [Edit]Dang! Vela must type faster than I do.
     
  5. Feb 5, 2013 #4

    xio

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    Vela, LCKurtz,

    so I map my xy-triangle onto the uv-plane:

    uv_map.png

    and evaluate the integral accordingly: $$\iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS=-2\iint_{T}\, du\, dv=-2\int_{0}^{0.5}\left[\int_{-u}^{u}dv\right]du=-\frac{1}{2}. $$ However, as you may see I get a negative value. Is this what is meant by "make a decision on the correct orientation"? What does it mean in particular?
     
    Last edited: Feb 5, 2013
  6. Feb 5, 2013 #5

    xio

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    Hrm... the book says additionally: "Let ##\mathbf{n}## denote the unit normal to ##S## having a nonnegative ##z##-component", but I cannot see how it can help right now...
     
  7. Feb 5, 2013 #6

    xio

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    So there can be two normals to the surface, each in the opposite direction: $$\mathbf{n}_{1}=\frac{\partial\mathbf{r}}{ \partial u}\times\frac{\partial\mathbf{r}}{\partial v}/\left\Vert \frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\Vert ,\mathbf{n}_{2}=-\mathbf{n}_{1}. $$ Since the area cannot be negative we simply make an arbitrary decision concerning the direction of the normal (take ##\mathbf{n}_2## in this case), is this correct?
     
    Last edited: Feb 5, 2013
  8. Feb 5, 2013 #7

    vela

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    Which normal has the non-negative z component?
     
  9. Feb 5, 2013 #8

    xio

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    In our case $$\mathbf{n}=\begin{vmatrix}\boldsymbol{\hat{ \imath}} & \boldsymbol{\hat{\jmath}} & \boldsymbol{\hat{k}}\\ 1 & 1 & -2\\ 1 & -1 & 0
    \end{vmatrix}=(-2,-2,-2) , $$ but since its ##z##-component is negative we take ##-\mathbf{n}=(2,2,2)##; is this correct?
     
  10. Feb 5, 2013 #9

    vela

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    Yes, that's correct.
     
  11. Feb 5, 2013 #10

    xio

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    Hurray!

    Vela, LCKurtz, thanks a lot!
     
  12. Feb 5, 2013 #11

    LCKurtz

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    You're welcome. I don't know why textbooks don't write the formula as$$
    \iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS=\pm \iint_{T}\mathbf{F}[\mathbf{r}(u,v)]\cdot\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\, du\, dv$$with the sign chosen to agree with the orientation given by ##\vec n##.
     
  13. Feb 6, 2013 #12

    xio

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    If fact, at least one does -- Apostol's "Calculus: II", page 434, section 12.9 "Other notations for surface integrals": first the relation between the direction of the normal and its sign is discussed and then the following formula is introduced: $$\begin{eqnarray*}
    \iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS & = & \iint_{T}\mathbf{F}[\mathbf{r}(u,v)]\cdot\mathbf{n}(u,v)\,\left\Vert \frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right\Vert \, du\, dv\\
    & = & \pm\iint_{T}\mathbf{F}[\mathbf{r}(u,v)]\cdot\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\, du\, dv.
    \end{eqnarray*}$$ (AFAIU since there's no ##\mathbf{n}## in the third expression, the "information" about the sign of the normal should be carried explicitly.)

    But since it was made explicit only in an "Other notations" section and, perhaps, since I didn't have a full understanding of the surface integrals, not taking it into account properly became the source of a minor confusion.
     
    Last edited: Feb 6, 2013
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