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I'm havin problems understanding the integration law by Gauss which states:

$$\iiint\limits_{G} \operatorname{div}(\vec{w})\, dV = \iint\limits_{\partial G} \vec{w} \cdot \vec{n } \, dA$$ (I don't know how to make a double closed integral in latex).

The main confusing thing is the change of the dA in dV, for example:

Let $B = \{x\in\mathbb{R}\, |\, x_1^{2}+x_2^{2}+x_3^{2} \leq 1\}$ with vectorfield $\vec{w}(x) = (x_1^{3},x_2^{3},x_3^{3})$.

With Gauss we know see that:

$$\iiint\limits_{G} \operatorname{div}(\vec{w})\, dV = \iint\limits_{\partial G} \vec{w} \cdot \vec{n } \, dA,$$

$$\text{div}(\vec{w}) = 3(x_1^{2}+x_2^{2}+x_3^{2})$$

so using spherical coordinates we get:

$$\iiint\limits_{G} \operatorname{div} (\vec{w}) \, dV = 3 \iiint\limits_{G} r^{2}\, dV.$$

Now I thought that $dV = d\phi d\theta dr$ but if i see my answers module it's supposed to be $dV = r^{2}\sin{\theta}d\phi\, d\theta \, dr$. My question now is why is this true(especially where did the sinus come from)?

Kees

P.s: $\vec{n}$ is a normal vector which points outside/away from the sphere.

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3 Answers 3

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Whenever you make a coordinate change you need to consider how volume will change. In this case you are using spherical coordinates, therefore the usual substitution is

$$ \begin{align} x & = r \cos \theta \sin \varphi, \\ y & = r \sin \theta \sin \varphi, \\ z & = r \cos \varphi, \end{align} $$

with limits $0 \leq \theta \leq 2 \pi$ and $0 \leq \varphi \leq \pi$. The "volume element" gets changed according to the following:

$$dV = dx \,dy \, dz = \left\vert \frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)} \right\vert \, dr \, d \theta \, d \varphi.$$

The number

$$\frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)}$$

is called the Jacobian, it represents the determinant of the following matrix, called the Jacobian matrix:

$$ \frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)} = \det \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \varphi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \varphi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \varphi} \end{vmatrix}.$$

If you compute this you will find $ - r^2 \sin \varphi$. The negative sign means that this coordinate change performs an orientation reversal, but since we are interested in volume per se we take the absolute value.

Therefore, your triple integral becomes

$$\iiint\limits_{G} r^2 \, dV = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\pi} \hspace{-5pt} \int_0^1 r^2 \cdot r^2 \sin \varphi \, dr \, d \varphi \, d \theta.$$

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The r^2 is there because the farther away from the origin you go, the larger a volume the same dr, dϕ and dθ will define. This volume will also depend on θ; sinθ is the distance to the z-axis, that is the radius of the circle a change in ϕ "moves along".

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hm..... I found out i forget the jacobian which is required when using substitution rules on integrals in more dimensions, i don't quite understand your explanation yet though, is it easier to understand when i draw a graph of the situation? –  Kees Til Feb 6 at 15:08
1  
Yes the r2sinθ factor is indeed the jacobian determinant of the transformation to spherical coordinates. I tried a more intuitive explanation. If you draw or imagine the situation, you will see that the length of the curve that dϕ describes is proportional to sin θ, where as both thr lengths of the curves that are described by dθ and by dϕ are proportional to r. So the volume element is (rsin θ*dϕ)*(rdθ)*(dr). (Try to see it in your head, took me a good while back in the days :) –  Pifagor Feb 6 at 15:17

You can find the volume and area elements in arbitrary coordinates just by finding the coordinate basis vectors. In most coordinate systems, these basis vectors are not unit vectors.

For example, the basis vector $\vec e_\theta$ associated with the angle from the $z$-axis is given by

$$\vec e_\theta \equiv \frac{\partial \vec r}{\partial \theta} = r \hat \theta$$

Why is it not unit? Well, imagine a curve of constant $r$ and $\phi$. As $\theta$ goes from $0$ to $\pi$, you cover a distance of $\pi r$. At larger and larger distances, this basis vector gets longer and longer.

Similarly, the basis vector $\vec e_\phi = r \sin \theta \, \hat \phi$. Finally, given that $\vec e_r = \hat r$, you can derive the volume element like so:

$$dV = \vec e_r \cdot (\vec e_\theta \times \vec e_\phi) \, dr \, d\theta \, d\phi$$

This is equivalent to using the Jacobian, but the scalar triple product emphasizes that we're looking at the volume of a parallelepiped.

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