Integrating a function over a rectangle

Question

I was just wondering why we integrate functions over shapes such as rectactcles, or circles.

So, if we integrate a function between two points sit represents its area. But if we integrate a function over a double integral, what does this represent?

Sorry if this seems a silly question, I'm just having trouble understanding it.

Answer 1

This represents volume. In general, integration increases the object being characterized by one dimension.

Answer 2

When you integrate a function of one variable, say $f(x)=2x+1$ from say $x=0$ to $x=2$, you get the area under the graph (a part of a triangle, in a sense, in this case). In this example, that area is

$\int_0^2 \! 2x+1 \, \mathrm{d}x=6$.

You can say that we integrate $f(x)$ over a interval, namely $[0,2]$.

When you then integrate a function of two variables, say $g(x,y)=x^2+y^2$, you have two variables that can change. If you were to plot $g(x,y)$, what would it look like? It would be a surface in three dimensions, since for each $x$ and $y$ (i.e a point in the xy-plane) you get a point $g(x,y)$ that can be plotted on the $z$-axis. (Use wolfram alpha to plot this yourself or look here.)

Now under this graph (i.e surface), you can choose a rectangle in the xy-plane, for example. Integrating over this rectangle will then mean that you let both $x$ and $y$ vary and thus get a double integral. For example

$\iint\limits_D \, x^2+y^2 \mathrm{d} x\,\mathrm{d} y$

where we integrate over the rectangle we just call $D$. Of course we need to specify the limits for the variables, for example $-0.5\leq x \leq 0.5$ and $-0.5 \leq y \leq 0.5$. In this case the double integral above would be $\frac{1}{6}$.

This integral will be the volume of the 'stuff' (or volume under the graph) between the surface (i.e the graph of g) and the xy-plane in the same way as the simple-integral represented the area between the line (i.e graph of f) and the interval [0,2].

I hope this makes things clearer.