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Question

empty · Answer

An overly complicated way to solving integrals. Let's do this one $$\int_0^2 x^2 dx $$

Let's call our interval $D=[0,2]$ so that we can write our integral in a more complicated way and let's also write our scalar function here as the divergence of some vector:

$$\int_D \nabla \cdot \left( \frac{x^3}{3} \hat i \right) dx$$

Just to reassure you this is the same, calculating:

$$\nabla \cdot \left( \frac{x^3}{3} \hat i \right)= \frac{3x^2}{3} \hat i \cdot \hat i = x^2$$

Ok now let's just say this looks an awful lot like Stokes' theorem, so let's go ahead and change it to an integral over the boundary:

$$\int_D \nabla \cdot \left( \frac{x^3}{3} \hat i \right) dx = \int_{\partial D} (\hat n \cdot \frac{x^3}{3} \hat i)  \ dx^0$$

So wait what's the boundary of D? (that's what this funny symbol means the boundary of D, $\partial D$ ) Well since $D=[0,2]$ like I said earlier, the boundary is just the two end points, 0 and 2.

What's the unit normal vector to the interval at these points? Well we will pick the convention of outwards facing normals, so that means at 0 $\hat n = -\hat i$ and at 2 the unit normal will be $\hat n = \hat i$. We can draw a number line if you have doubts on this.

So now all we have to do is evaluate this integral at the endpoints, but there's only two so we add them up:

$$\int_{\partial D} (\hat n \cdot \frac{x^3}{3} \hat i)  \ dx^0 = -\hat i \cdot \frac{0^3}{3} \hat i + \hat i \frac{2^3}{3} \hat i$$

Ok it looks like we're getting:

$$\int_0^2 x^2 dx = \frac{8}{3}$$

So what do you think?

irishboy123 · Answer

hey Empty

so Stokes takes a curl over an area and turns it into a line integral around a closed loop.

here, currently, i see a div.  am i missing something?!

empty · Answer

Ahh well you can derive Stokes theorem from the divergence theorem so I suppose I should have said divergence theorem whoops, oh well.

irishboy123 · Answer

lol!