Fundamental theorem of calculus for double integral

[Jhenrique]

The popular fundamental theorem of calculus states that \int_{x_0}^{x_1} \frac{df}{dx}(x)dx = f(x_1)-f(x_0) But I never see this theorem for a dobule integral... The FTC for a univariate function, y'=f'(x), computes the area between f'(x) and the x-axis, delimited by (x₀, x₁), but given a bivariate function, z_xy=f_xy(x,y), you can want to compute the volume between a surface f_xy(x,y) and the xy-plane, delimited by ((x₀, x₁), (y₀, y₁)). So, this volume is given by double integral \iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f}{\partial y \partial x}(x,y)dxdy = f(x_1,y_1)-f(x_0,y_0) Correct? If yes, what happens that anyone mentions this!?

 

[D H]

Your notation doesn't make a bit of sense. A double integration is over an area, not from one point to another.

The multi-dimensional analog of the fundamental theorem of calculus is Stoke's theorem.

 

[Jhenrique]

Your notation doesn't make a bit of sense. A double integration is over an area, not from one point to another.

The multi-dimensional analog of the fundamental theorem of calculus is Stoke's theorem.

Really? But if a simple integral can compute areas so a double integral can to compute volumes. Also, the stokes theorem conffirm that volumes can be computed by double or triple integrals.

If the double integral of the 1st post not compute this volume:

So, what can computeit?

 

[PeroK]

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f}{\partial y \partial x}(x,y)dxdy = f(x_1,y_1)-f(x_0,y_0) Correct? If yes, what happens that anyone mentions this!?

Try:
f(x, y) = x^2 y^2
If you evaluate the double integral this will show that your hypothesis is false.

 

[D H]

Try:
f(x, y) = x^2 y^2
If you evaluate the double integral this will show that your hypothesis is false.

I'd say it's worse than false. It's not even wrong.

 

[csmallw]

Jhenrique, what the others aren't bothering to tell you is that your hypothesis is following the right line of intuition. The correct generalization of the fundamental theorem of Calculus to a two-dimensional integration across a rectangular region is given by

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

I should stress that my formula only applies if the region of integration is rectangular, and if it is orientated along the Cartesian coordinates (and perhaps if a few other assumptions like "smoothness" and "continuity" are satisfied--I'm a physicist, not a mathematician). The specificity of the equation is probably why you do not often see it highlighted by teachers. More generally, you can perform 2D integration when x0 and x1 are functions of y, but the above form is no longer accurate or useful. I think that techniques in multidimensional integration are usually explained pretty well in most calculus books. If you haven't already done so, you should check out, for example, "Multivariable Calculus" by James Stewart.

Also, setting D H's derisive sense of humor aside, I do recommend that you take the advice of reading up a bit on Stoke's theorem. The above formula should merely be a special case of that more general statement.

 

[Jhenrique]

Yeah, the ideia is this!

But, I not found anything similar to your equation:

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

How you came to this result?

 

[PeroK]

To prove:

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

You just have to integrate twice:

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = \int \limits_{y_0}^{y_1} \frac{\partial f(x_1,y)}{\partial y} - \frac{\partial f(x_0,y)}{\partial y}dy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

You should have tried with the function I suggested: f(x, y) = x^2y^2

You can't prove a formula by examples, but if you try a specific function it might give you an idea of what is happening. Also, one counterexample is all you need to show a formula is not true.

 

[Jhenrique]

To prove:

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

You just have to integrate twice:

\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = \int \limits_{y_0}^{y_1} \frac{\partial f(x_1,y)}{\partial y} - \frac{\partial f(x_0,y)}{\partial y}dy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0).

You should have tried with the function I suggested: f(x, y) = x^2y^2

You can't prove a formula by examples, but if you try a specific function it might give you an idea of what is happening. Also, one counterexample is all you need to show a formula is not true.

ROX!

But if this integral $\int_{x_0}^{x_1} \frac{df}{dx}(x)dx = f(x_1)-f(x_0)$ is analogous to this vectorial integration $\int_{\vec{s}_0}^{\vec{s}_1}\vec{\nabla}f\cdot d\vec{s}=f(\vec{s}_1)-f(\vec{s}_0)$, so which is the vectorial integration analogous to $\iint \limits_{y_0 x_0}^{y_1 x_1} \frac{\partial^2 f(x,y)}{\partial y \partial x}dxdy = f(x_1,y_1)-f(x_1,y_0)-f(x_0,y_1)+f(x_0,y_0)$ ?