Easy question : $\int (xdy+ydx)$

Question

I am ashamed to ask such an easy question but, well: Lets say I got a function $$ f(x,y)=xy $$ Now let's compute the total differential of the function $$ d(f(x,y))=xdy+ydx $$ Now if I do $$ \int d(f(x,y))=\int(xdy+ydx)=\int xdy +\int ydx =x\int dy +y \int dx = 2xy \neq f(x,y) $$ Where is the bias in the reasoning ? Thanks in advance

Answer 1

You're being sloppy in defining the integral.

When you write $\int \mathrm d(f(x,y))$ you are really writing a line integral which needs to have some path specified. Since you are integrating an exact differential the value of this line integral will equal the difference in the value of the function at the beginning and end points of the path.

For instance we could let the path be the straight line which connects the origin to the point $(x_0,y_0)$. If we call this path $\gamma : [0,1] \rightarrow \mathbb{R}^2$ then we can write $\gamma(t) = \langle x_0 t , y_0 t \rangle$ and the line integral becomes,

$$\begin{align*} \int_{C} ydx + xdy &= \int_0^1 y(t) \frac{dx}{dt} dt + x(t) \frac{dy} {dt}dt \\&= \int_{0}^1 (y_0 t) x_0 dt + (x_0t) y_0 dt \\&= x_0 y_0 \int_{0}^1 2 t dt \\&= x_0 y_0 \left[ t^2 \right]_0^1 \\&= x_0 y_0 \end{align*}$$

Answer 2

Properly $\int x \ dy + y \ dx $ is a path integral, so your treatment is not correct as a path is not specified.

For example, suppose we start at $(x_1,y_1)$ and end at $(x_2,y_2)$ in two line segments, first parallel to the $x$ axis $L_1 : (x_1,y_1) \to (x_2,y_1)$ and then parallel to the $y$ axis $L_2 : (x_2,y_1) \to (x_2,y_2)$.

Then

$$\int_{L_1 + L_2} x \ dy + y \ dx = \int_{L_1 + L_2} d(xy) = [ xy ]_{(x_1,y_1)}^{(x_2,y_2)} = x_2y_2 - x_1y_1$$

And piecewise $$\begin{align} \int_{L_1 + L_2} x \ dy + y \ dx & = \int_{L_1} x \ dy + \int_{L_1} y \ dx + \int_{L_2} x \ dy + \int_{L_2} y \ dx \\ & = \int_{y_1}^{y_1} x \ dy + \int_{x_1}^{x_2} y_1 \ dx + \int_{y_1}^{y_2} x_2 \ dy + \int_{x_2}^{x_2} y \ dx \\ & = \quad 0 \quad\ \ \ \ + y_1(x_2-x_1) + x_2(y_2-y_1) \ + \quad 0 \\ & = x_2y_2 - x_1y_1 \end{align}$$ ...which is consistent.

Answer 3

Thanks a lot for your answer. One last question, the way to identify that it is a line integral is this ? $$ \dfrac{\partial^2 f}{\partial_x \partial_y}=\dfrac{\partial^2 f}{\partial_y \partial_x}\neq0 $$ And if I would have $$ \dfrac{\partial^2 f}{\partial_x \partial_y}=\dfrac{\partial^2 f}{\partial_y \partial_x}=0 $$ It would be valid to simply do $$ \int (f(x,y)) $$ Or I did not get it right?