Evaluating a surface integral using parametric/explicit representation

[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).

 

[vela]

[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.

 

[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.

[Edit]Dang! Vela must type faster than I do.

 

[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?

 

[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...

 

[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?

 

[vela]

Which normal has the non-negative z component?

 

[xio]

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?

 

[vela]

Yes, that's correct.

 

[xio]

Hurray!

Vela, LCKurtz, thanks a lot!

 

[LCKurtz]

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 $\vec n$.

 

[xio]

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.