# Surface Area Double Integral

1. Jan 11, 2005

### Cyrus

Hi,

I have a question on the method of calculation of the surface area of a surface. I am using "Calculus Concepts and contexts by stewart", chapter 12.6.

In it, he goes on to explain how to calculate the suface area of a surface as a double integral by using approximations. He breaks up the projection of the surface onto the x-y plane into a grid. And each grid maps out a grid onto the surface as well. Now he wants to find an approximate area of each of those squares, and sum them. A rieman sum basically. Now on the projection he lables each side of the square delta x, and delta y.

He then goes on to say that he needs to approximate each of those patches on the surface, and that can be done if you use two vectors that start at one corner of the patch and both end at each end of the patch. You can then do the cross product of these two vectors to find the area of one patch. Similarly, you can do this for all the patches and sum it.

Now, in order to approximate these vectors, he does the partial with respect to x, to find the tangent to approximate one side of the patch, and similarly with respect to y for the other side. These gives him two tangent vectors that lie on the tangen plane. And he crosses them to approximate the area of the patch.

Here come the confusion. When he calculates the area, he multiplies the tangent with respect to x, by delta x, and similarly for y, by delta y. Why does he multiply the tangents by these values?

Also, he WANTS the two vectors to approximate the sides of the patch, so that the area closely computes the area of the patch; however, the tangent vectors WRT x, and WRT y, can have different magnitudes depending on the origional function that defines the surface. So what makes them work out so that the magnitude will ALWAYS give a good approximation for the area of the patch?

If the patch is very small, then the tangent vectors WRT x and y might be way too big. So when you cross them, you get a huge area relative to what the patch area SHOULD be. So what fixes this dilema?

Cyrus

2. Jan 12, 2005

### Galileo

Consider a small patch of surface traced out by the vector $\vec r(u,v)$ when u goes from $u_i$ to $u_i+\Delta u$ and v from $v_i$ to $v_i+\Delta v$.
Then the patch can ve approximated by the parallelogram spanned by the vectors $\vec r(u_i+\Delta u_i,v_i)-\vec r(u_i,v_i)$ and $\vec r(u_i,v_i + \Delta v)-\vec r(u_i,v_i)$.
These in turn are approximated by $\vec r_u(u_i,v_i)\Delta u$ and $\vec r_v(u_i,v_i)\Delta v$.
Because clearly:
$$\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)$$.

That's why the surface area patch is approximated by:
$$|(\Delta u \vec r_u(u_i,v_i) \, \times \, \Delta v \vec r_v(u_i,v_i)|=|\vec r_u(u_i,v_i) \times \vec r_v(u_i,v_i)|\Delta u \Delta v$$.

Note that it is NOT the projection of the surface onto the xy-plane that is cut into a grids, but the domain D of the function, which is some region in the uv-plane.

For the question about the tangent vectors. It is true that sometimes they may be very big, but that is because of the partical parametrization of the surface that is chosen. Recall that $\vec r_u$ can be considered as the speed at which a curve in the surface is traced out. If the speed is big a greater side of the parallelogram we considered is traced out when u goes a little distance $\Delta u$, so the surface area patch will be greater as well.
You may be more off from the true area of the surface patch when $r_u$ is big, but in the limit as $\Delta u$ goes to zero it approximation becomes exact. Fortunately it can be proven that the surface area is independant from the chosen parametrization.

(The same thing occurs in a standard one dimensional integral. If f(x) varies wildly (almost vertically, thus f'(x) very big), the sum of rectangles approximation gets bad, but in the limit $\Delta x \to 0$ it will still be exact).

Last edited: Jan 12, 2005
3. Jan 12, 2005

### Cyrus

But this: $\vec r(u,v)$ is for a scalar function. How about when it is a vector function: $\vec r(u,v)=x(u,v)i + y(u,v)j + z(u,v)k$

What meaning does the partial have in this case, becuase it is now a vector function.

4. Jan 12, 2005

### Galileo

I did mean for $\vec r(u,v)$ to be a vector-valued function (hence the arrow on the $\vec r)$.

Suppose you hold v constant and vary u. Then $\vec r(u,v)$ traces out a curve in $\mathbb{R}^3$.
$\vec r_u(u,v)$ simply gives the tangent vector along this curve, just like if you where considering a vectorfunction of one variable $\vec r(t)$. (It's the rate at which the curve is traced out).
If you vary v and hold u constant, you trace out a different curve ofcourse and $\vec r_v$ is interpreted similarly.

5. Jan 12, 2005

### Cyrus

A question on how you used the following:

You see, I have seen this notation used on a scalar function but not on a vector function.

6. Jan 12, 2005

### dextercioby

It doesn't really matter whether u meet scalar derivatives,or vector derivatives.Since scalars are vector components in a basis,thing are quite similar,since any vector function's derivative can,intur,be decomposed into scalar components.

Daniel.

PS.Think about physical situations.The simplest:velocity vector is the time derivative of the position vector.

Last edited: Jan 13, 2005
7. Jan 12, 2005

### Cyrus

Ok dex, heres a reworded question. I was not clear becuase I did not understand my question very well myself. Heres what I want to ask.

$$\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)$$

We have all this junk up above. Now lets say that we do the partial dervative of a parametric curve

$$r(u,v)=x(u,v)i + y(u,v)j+z(u,v)k$$

Now lets say its with respect to u, so that v is held fixed.

Then the partial derivative, (depending on the surface), may have a variable 'u' to some power in it for either the i,j, k or even all directions. I thought that the goal was to find a tangent vector that only had a component in two orthogonal directions. For instance, we want to find a tangent so that it goes delta x in the i direction and delta y in the j direction, but does not change in the k direction. Then we can approximate the area of the patch. But if you look above, it might have the variable 'u' in all three directions, so it might have a change in all three directions. Is this not a problem?

heres my point, one tangent vector should be parallel to the x axis, and the other to the y axis. But this isint true if the partial derivative contains a variable in each direction.

Last edited: Jan 12, 2005
8. Jan 13, 2005

### Cyrus

Take this for example,

$$\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)$$

r is a vector function. So would I have to do the following?

$$\frac{ x(u_i+\Delta u,v_i)+ x(u_i,v_i)}{\Delta u}i+ \frac{ y(u_i+\Delta u,v_i)-y(u_i,v_i)}{\Delta u}j + \frac{z(u_i+\Delta u,v_i)-z(u_i,v_i)}{\Delta u}k \approx \vec r_u(u_i,v_i)$$

Also, a level curve means that it is horizontal or verticle. But to clearify, this means that the curve is horiztonal or verticle on the domain, right? If you hold u fixed, and vary v, then it may or may not be horizontal or verticel on the SURFACE. That would depenend on how the surface is defined. This sort of anwsers my own question previously, but i dont know if im wright or not.

Last edited: Jan 13, 2005
9. Jan 13, 2005

### Galileo

No. It seems like you think the surface is approximated by horizontal patches. This is not the case. It is approximated by the parallelogram spanned by two tangent vectors. One in the direction of $r_u$ and the other in the direction of $r_v$. The surface patches could even by vertical. Like in the following case of a hollow cylinder of height h and radius R:

$$x(u,v)=R\cos(u),\, y(u,v)=R\sin(u), \, z(u,v)=v$$
with $0\leq u < 2\pi$ and $0 \leq v \leq h$.

Then since:
$$\vec r_u(u,v)=\frac{\partial x}{\partial u}\vec i+\frac{\partial y}{\partial u}\vec j+\frac{\partial z}{\partial u}\vec k=-R\sin(u)\vec i+R\cos(u)\vec j$$

and

$$\vec r_v(u,v)=\frac{\partial x}{\partial v}\vec i+\frac{\partial y}{\partial v}\vec j+\frac{\partial z}{\partial v}\vec k=\vec k$$
we see that $\vec r_v$ points straight up and a surface patch has an area of approximately:
$$|\vec r_u \times \vec r_v|\Delta u \Delta v = R\Delta u \Delta v$$

The tangent vectors can have ANY direction whatsoever. Since the surface can be anything at all. Like a sphere. The tangent vectors point in different directions all over the sphere.

If you have the same version of Stewart as I have. Check out paragraph 10.5 about parametric surfaces (Figure 5).

Last edited: Jan 13, 2005
10. Jan 13, 2005

### Cyrus

Ok, just to make sure though, the curve on the domain D; however, will be parallel to the x axis or y axis, if you take the partial derivative.

11. Jan 13, 2005

### Galileo

Sorry, I don't see the connection between the previous and level curves.

Level curves are defined for functions f of two variables.
They are the points in the domain of f for which f(x,y)=k. There's no connection with surface area.

12. Jan 13, 2005

### Galileo

There is no x- or y-axis for the domain D. Since D exist in the uv-plane.
There are two different sets at work here. $\vec r$ is a function from the uv-plane to $\mathbb{R}^3$, with x,y and z coordinates.

$$\vec r : D \to \mathbb{R}^3$$

13. Jan 13, 2005

### Cyrus

Ok, heres my point though, lets say you take the partial derivative of a function with respect to x, then is the tangent vector you get parallel to the x-axis?

P.S. whoops I dident mean level curves, I ment the curve you get when you hold on variable fixed and let the other variable vary, I cant think of what its called at the moment, it slipped my mind.

14. Jan 13, 2005

### Galileo

Plenty of room for confusion here. When you say the partial derivative of a function wrt x, then I assume you are talking about a function of two (or more) variables.
Set z=f(x,y). Then the partial derivative wrt x is not a vector, but a function of x and y.

If you mean $\vec r(u,v)$ then you cannot take the partial derivative with respect to x, because it's a function of u and v.

In light of what you said before:
And by level curve you meant the curve traced out when u varies but v is held constant. Then yeah, in the uv-plane, v is constant and u varies, so you're taking a horizontal line in D. And the curve that is traced out are those values of $\vec r(u,v)$ for which u and v lie on the line in D. The curve that is traced out can be anything, and it doesn't have to be horizontal.

Take the example of the cylinder before. If we hold u constant (say u=0, so x=R and y=0), then when we vary v, you trace out a vertical line parallel to the z axis with x coordinate R and y coordinate 0.

15. Jan 13, 2005

### Cyrus

Yes, I see what you mean now Galileo So we take the partial derivative of the vector function $$\vec r (u,v)$$.

16. Jan 14, 2005

### Cyrus

Ok, a few questions:

First, when you take the partial derivative of a scalar function, you get a scalar value, and this value is the SLOPE of the tangent to the grid curve at that point.

Second, when you take the partial derivative of a parametric function, you get a vector in return. But this vector is the tangent vector.

But I thought the Definition of a partial derivative means that you must get a scalar value, because the derivative is a slope, but we got a vector. So how does it make sense to talk about the partial derivative of a parametric function of two variables.

17. Jan 14, 2005

### FulhamFan3

That's the definition if you take the partial of a scalar function. If you take the derivative of a vector you get a vector. It makes sense if you want to study the equation by isolating one variable.

18. Jan 14, 2005

### Cyrus

I see, but here is my only concern:

$$\frac{\vec r(u_i+\Delta u,v_i)-\vec r(u_i,v_i)}{\Delta u} \approx \vec r_u(u_i,v_i)$$

You see, if this were a scalar function, then it would mean, we approximate the value of the tangent line, which is the slope as x or y varies while the other is held fixed.

But now look at the parametric function, its clear that the partial derivative of u or v, on the surface is not a straight line, as galileo pointed out, it could be a curve. So the tangent now has components in x,y,and/or z. So its no longer meaningful to talk about the slope anymore. Is the components of the partial derivative the slope in the i,j and k directions?

19. Jan 15, 2005

### Galileo

Per definition:
$$\vec r_u(u,v)=\lim_{\Delta u \to 0}\frac{\vec r(u+\Delta u,v)-\vec r(u,v)}{\Delta u}$$
It tells you how fast the vector $\vec r$ is changing in the point $r(u,v)$ as you vary u. It's very similar to the vector function in 1D, where $\frac{d}{dt}\vec r(t)$ gives you the speed at which $\vec r$ traces out the curve at the point $\vec r(t)$.
So the derivatives of the vectors do NOT give the slope of the surface.

20. Jan 15, 2005

### Cyrus

Hmm, lets see if this is right or not.

In the case of x=x, y=y and z=f(x,y), when you do the partial derivative you get a scalar, (the slope).

When you do r(u,v) you get a vector.

Is the "slope" of the scalar case, equal to the magnitude of the vector in the r(u,v) case? I hope this is clear, is the scalar a special case of the parametric form, (since it has only two components to the tangent, for example a change in x and z, or y and z.).

Hmm, let me state it like this. Lets say I wrote a function as r(x,y)= xi+yj+f(x,y)k.
Now theres two ways I can look at this right? If I just do the partial derivatve of $$f_x (x,y)$$ I will get some scalar value that is the slope, right? But lets say I also do the partial derivative of this vector $$r_x(x,y)$$ Now I DONT get a scalar slope, but a vector tangent, right? But what if I found the MAGNITUDE of this tangent, would I get the same anwser as the scalar slope of $$f_x(x,y)$$? I hope this example made my question clear for you.

Edit: after looking it over I dont think what I said is correct; however, can it be interpreted in this way, when i do $$f_x(x,y)$$ I get the SLOPE of the tangent vector, but I dont know what the vector is, where it points, or what is magnitude is. When I do $$r_x (x,y)$$ I know the tangent vector at that point, I know its components in the i,j and k directions, and I know the magnitude of the tangent vector, but I know nothing of its slope.

Last edited: Jan 15, 2005