Evaluating a surface integral using parametric/explicit representation

In summary: Parametric representation:Given ##\mathbf{r}(u,v)=(u+v,u-v,1-2u) ##, we find $$\iint_{S}\mathbf{F}\cdot\mathbf{n}\, dS=\iint_{T}(u+v,u-v,1-2u)\cdot(-2,-2,-2)\, du\, dv=-2\iint_{T}\, du\, dv. $$
  • #1
xio
11
0
[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).
 
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  • #2
xio said:
[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## .
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.
 
  • #3
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.
 
  • #4
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?
 
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  • #5
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...
 
  • #6
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?
 
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  • #7
Which normal has the non-negative z component?
 
  • #8
vela said:
Which normal has the non-negative z component?

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?
 
  • #9
Yes, that's correct.
 
  • #10
Hurray!

Vela, LCKurtz, thanks a lot!
 
  • #11
xio said:
Hurray!

Vela, LCKurtz, thanks a lot!

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##.
 
  • #12
LCKurtz said:
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##.

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.
 
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FAQ: Evaluating a surface integral using parametric/explicit representation

1. What is a surface integral?

A surface integral is a mathematical tool used to calculate the area or volume of a three-dimensional surface. It is a generalization of the concept of a line integral, which calculates the length of a curve in two dimensions.

2. What is the difference between parametric and explicit representation?

In parametric representation, a surface is described by a set of equations in terms of one or more parameters. These equations define the coordinates of points on the surface. In explicit representation, the surface is described by a single equation in terms of x, y, and z coordinates.

3. How do you evaluate a surface integral using parametric representation?

To evaluate a surface integral using parametric representation, you first need to determine the parametric equations that define the surface. Then, you can use these equations to calculate the limits of integration and the integrand. Finally, you can solve the integral using standard integration techniques.

4. How do you evaluate a surface integral using explicit representation?

To evaluate a surface integral using explicit representation, you first need to determine the equation that defines the surface. Then, you can use this equation to calculate the limits of integration and the integrand. Finally, you can solve the integral using standard integration techniques.

5. What are some real-world applications of surface integrals?

Surface integrals have many applications in physics and engineering. They are used to calculate the mass, center of mass, and moment of inertia of objects with complex shapes. They are also used in fluid mechanics to calculate the flow of a fluid over a surface. Additionally, surface integrals are used in computer graphics to render 3D objects and in thermodynamics to calculate the heat transfer across a surface.

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