# Evaluating a surface integral using parametric/explicit representation

1. Feb 4, 2013

### xio

[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).

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
2. Feb 4, 2013

### vela

Staff Emeritus
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. Feb 4, 2013

### LCKurtz

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.

Dang! Vela must type faster than I do.

4. Feb 5, 2013

### xio

Vela, LCKurtz,

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

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
5. Feb 5, 2013

### xio

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. Feb 5, 2013

### xio

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
7. Feb 5, 2013

### vela

Staff Emeritus
Which normal has the non-negative z component?

8. Feb 5, 2013

### xio

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. Feb 5, 2013

### vela

Staff Emeritus
Yes, that's correct.

10. Feb 5, 2013

### xio

Hurray!

Vela, LCKurtz, thanks a lot!

11. Feb 5, 2013

### LCKurtz

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. Feb 6, 2013

### xio

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