# Thread: Green's theorem and intuition

1. ## Green's theorem and intuition

When I see a double integral, I always get a thought that we are calculating the volume under the graph in 3D coordinates. Like we are calculating the area for single integral.

But this is not the case in an example of Green's theorem. Or other similiar theorems. How can I intuitive remember why is there a double integral in Green's theorem and what are we calculating?

Thanks.

2. ## Re: Green's theorem and intuition

Originally Posted by Nforce
When I see a double integral, I always get a thought that we are calculating the volume under the graph in 3D coordinates. Like we are calculating the area for single integral.

But this is not the case in an example of Green's theorem. Or other similiar theorems. How can I intuitive remember why is there a double integral in Green's theorem and what are we calculating?

Thanks.
I would suggest you need to broaden your perspective a bit. It is in fact more natural for double integrals to be used for area related things rather than volumes, although that is one use for them. For example consider a thin plate represented by an area $P$ in the $xy$ plane. If you do $\iint_P ~1~dA$ you get the area of the plate. If the plate had an area density $\delta$ in units of mass/area the integral $\iint_P~\delta~dA$ gives the mass of the plate. And you have the usual integrals giving the first and second moments and center of masses. They are all naturally worked with double integrals and they have nothing to do with volumes. Sometimes double and triple integrals are related by some physical property. You haven't indicated what level of calculus you are studying, but an example is Stoke's Theorem which relates a triple and double integral. The idea that a flux flowing through a closed surface (a double integral) can be calculated by the divergence of the field over the enclosed volume (a triple integral). Green's theorem is like that. It relates the value of a line integral around an enclosing curve to a double integral over the enclosed region. With the right choice of $P$ and $Q$, you can calculate the area by integrating around the boundary. If you think about it, that makes sense, because the bounding curve certainly determines the area.

3. ## Re: Green's theorem and intuition

Sorry but I still don't know. It seems reasonable for $\displaystyle dA = dxdy$ we have an infinitesimally small area which we sum up, and we get a large area. So if we have a function which is $\displaystyle f(x,y)$ then we get a volume under it, but not in the case of an area? The area is defined by what in mathematical problems?

4. ## Re: Green's theorem and intuition

Originally Posted by Nforce
Sorry but I still don't know. It seems reasonable for $\displaystyle dA = dxdy$ we have an infinitesimally small area which we sum up, and we get a large area. So if we have a function which is $\displaystyle f(x,y)$ then we get a volume under it, but not in the case of an area? The area is defined by what in mathematical problems?
Summing the dxdy elements is the same as doing the integral $\iint_A~1~dydx$ which gives the area. It also gives the volume of a solid of height $1$ over the area because the volume of a solid of constant height $1$ is $1 \times \textrm{area of base}$ which is the same number.

6. ## Re: Green's theorem and intuition

Originally Posted by Nforce
I don't know. What about it? Its integral can be interpreted as a volume, but that is only one interpretation, and in many problems that isn't a helpful interpretation. Did you understand post #2?

7. ## Re: Green's theorem and intuition

Yes, I think I understand your post #2.

So for function f(x,y), its integral can be interpreted as a volume. OK, but what are the other interpretations?
If I understand correctly it's all how do we interpret the integral.

8. ## Re: Green's theorem and intuition

Originally Posted by Nforce
Yes, I think I understand your post #2.

So for function f(x,y), its integral can be interpreted as a volume. OK, but what are the other interpretations?
If I understand correctly it's all how do we interpret the integral.
I thought I gave you several examples in post #2. I'll give it one more try. The interpretation of an integral depends on what you are calculating with it. Think back to your first calculus course when you first did integrals. You were likely working problems like "Find the area under the curve $y = f(x)$ for $a\le x \le b$, where $f(x) \ge 0$ on $[a,b]$". You learned to do this by calculating $A =\int_a^b~f(x)~dx$. So integrals like that can be interpreted as the areas under curves. But what if you have an object whose velocity at time $t$ on the $x$ axis is given by $v = f(t)$ where $f(t) \ge 0$ on $[a,b]$ and you want to know how far the object travels from $t = a$ to $t = b$? Since postition is the antiderivative of velocity, you can calculate $\textrm{Distance s = change in position = }s(b)-s(a) = \int_a^b~f(t)~dt$ by the fundamental theorem of calculus.

So here we have an integral $\int_a^b~f(t)~dt$ which is just like $\int_a^b~f(x)~dx$. But we aren't calculating area. We are calculating distance travelled. And while it is true that $\int_a^b~f(t)~dt$ can be interpreted as the area under a curve, that isn't very helpful for this problem and is irrelevant to the problem. It's the same with double integrals. Maybe this will help you understand post #2 better.

To repeat: The interpretation of an integral depends on what you are calculating with it.