Difference between these two surface integrals?

[Xyius]

In my Calculus book, in the chapter that introduces multiple integration, it has a chapter on integrals that calculate the surface area of a function in space. They define the integral to be..

\int \int dS = \int \int \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA

However, one chapter later they have another chapter entitled "Surface Integrals" where they define the surface area of a function in space to be..

\int \int f(x,y,z) dS = \int \int f(x,y,g(x,y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA

What is different between these two integrals? They both say they calculate surface area.

EDIT:

From what I can gather, the first one is for an area right under the function in space, and the second one is for a region other than the base under the curve. Is this correct?

Thanks!
~Matt

 

[Jerbearrrrrr]

The second one looks like it's integrating the value of a function over a surface. It's hard to answer since we don't know what all the variables mean, or even what space we're working in (though I assume 2 or 3D space).

Wouldn't be able to take a picture of the textbook would you? Or if it's the Stewart's essential transcendental crocodiles or whatever that book is, I could go get it quick.
[edit]
Also, functions don't really have areas.
I know we've all been taught that the integral of y(x)dx is the area under a curve...
but you can't hold onto that way of thinking forever (even though you can generalize it somewhat for a while).

Shapes in space have areas associated with them.
Functions assign a number (or something like a number) to each point in space.
Well anyway, in my first vec calc stuff, I found it easier to think of integrals as sums.

 

[Xyius]

Cant take a picture at the moment, but I think you may be right. I think the first one Calculates the surface area of the actual curve in space, and the second one calculates the function values over another surface.

 

[Jerbearrrrrr]

What's this g(x,y) thing?

Oh. The surface is z=g(x,y)?
Not sure why the derivatives of 'f' appear in the sqrt, but it's early in the morning Dx

 

[LCKurtz]

In my Calculus book, in the chapter that introduces multiple integration, it has a chapter on integrals that calculate the surface area of a function in space. They define the integral to be..

\int \int dS = \int \int \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA

That is the area of as surface parameterized as z = f(x,y). The double integral is taken over the (x,y) domain of the function f, and the value of the integral is the surface area of the surface given by z = f(x,y).

However, one chapter later they have another chapter entitled "Surface Integrals" where they define the surface area of a function in space to be..

\int \int f(x,y,z) dS = \int \int f(x,y,g(x,y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA

What is different between these two integrals? They both say they calculate surface area.

Are you sure that one is copied right? More typical would be something like

\int \int \delta(x,y,z) dS = \int \int \delta(x,y,f(x,y)) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA

This would represent the mass of the same surface where its variable area density is \delta(x,y,z). If the density is 1 it gives the same formula for surface area as above.

EDIT:

From what I can gather, the first one is for an area right under the function in space,

It isn't the area "right under" the graph. It is the surface area of the surface itself.

 

[darkside00]

depends what your doing. are you looking at a single surface of a function?, or a function with a boundary over a surface?

 

[Xyius]

LCKurtz, what you say makes sense, and there are examples that find the mass of a surface with a variable density. I did copy it down incorrectly. I understand now. Thanks :D