Notation for double-integrals - partial or full differentials? When you are trying to find a volume for a function z = f(x,y), the common notation is to find:
$$\int\biggr(\int f(x,y)dx\biggr)dy$$
However, when you do this, you are actually keeping the $y$ constant on the first integral.  To me, for this to be the way it works, it seems like you should actually be using the partial differential $\partial x$ and $\partial y$.  So it seems like the notation for this should be:
$$\int\biggr(\int f(x,y)\partial_z{x}\biggr)\partial_z{y}$$
Is this an incorrect intuition?  Why or why not?
 A: Not quite. "Keeping $y$ constant" is not about the differential, but about the path of integration.
Recall that in single variable calculus, a definite integral might be written as
$$ \int_0^1 f'(x) \, \mathrm{d}x = f(1) - f(0) $$
That subscript and superscript are specifying a path that one is integrating over. In one dimension, calculus is fairly simple and you only need to specify where the path begins and ends; i.e. you integrate from $0$ to $1$. Any other details about the path turn out to be irrelevant.
In two dimensions, there are many more possibilities for paths. Fairly arbitrary paths are possible to integrate over; see line integral. (also called "path integral")
The partial integral that you are referring to is saying that the integral should be taken over a path of constant $y$. The differential is still going to be $\mathrm{d}x$.
A: There are some authors who use partial dif notation for multivariable intagrals and they do the right thing (for example Ross). But when you performing a double integral to two variable function with two variables of it, dxdy it self becomes for example dA and that is a exact dif of the complete function. This statement is true for 3 or more variable functions.
A: Apparently, you are correct, because when you use implicit differentation to solve partial derivatives of multivariable functions in the form $z = f(x, y)$, you will take the partial derivative of $z$ with respect to $x$, which means you will take the derivative of the $x$-terms and leave all $y$ terms as constants. It will be reversed when you take the partial derivative of $z$ with respect to $y$; you will treat all $x$-terms as constants and calculate the derivative of all the $y$-terms. Since it looks like it applies to integrals, yes, it looks like you would use $\partial x$ and $\partial y$.
