# How do you integrate magnitudes of vectors not just scalars?

I am working on a method of integrating over the of shapes, starting with a sphere to find its mass to find its gravitational pull with a uniform density in space. I know the volume of a sphere in general is normally given by

$\int_{0}^{ \pi} \int_{0}^{2 \pi} \int_{0}^{R}drd \theta d \phi$.

and Newton's law of gravity given by

$F=\frac{G M_1 \cdot M_2}{r^2}$

where $M_1$ is the mass of a the spherical object like a star or cloud and $M_2$ would allegedly be the mass of something within the sphere like a planet.

In actual 3D space though, simple scalars aren't enough and the force of gravity is defined between the specific locations of objects, meaning I have to deal with vectors. Since the density of uniform, I would think it acts like a constant, but there are still all kinds of directions from different particles, and combining the two equations for integrating density over volume would give mass

G$\int_{0}^{ \pi} \int_{0}^{2 \pi} \int_{0}^{R} \frac{ M_{2} \rho \hat{\mathbf{r}} }{\| \mathbf{r} \|^{2}}drd \theta d \phi$

The problem is, I don't see how to just spontaneously integrate a vector. I can't think that after 300 or 400 years that no one has figured out how to do it. I could maybe say $r = \sqrt{x^2+y^2+z^2}$ but then I would have to define $\theta$ and $\phi$ in Cartesian coordinates and I don't see a reason to bother with that when I'm already in spherical coordinates. As far as I know, $r$ is its own coordinate.

• According to your formula the volume of a sphere is $2\pi^2 R$. – Christian Blatter Mar 13 '18 at 5:56
• Right, so obviously I know this process is wrong, which is why I'm asking how it can be corrected. – John Joe Mar 13 '18 at 21:25

As for (i) you have to understand an integral in the following way: You are given a domain $B\subset {\mathbb R}^d$ and a real-, complex-, or vector-valued function $f:\>B\to{\mathbb X}$ describing a certain economical, geometrical, or physical quantity depending on and changing with $x\in B$. The integral $$\Phi(B, f)=\int_B f(x)\>{\rm d}(x)$$ then captures the "total impact" realized by this $f$ on $B$. From intuitively obvious additivity and linearity considerations it then follows that $\Phi(B,f)$ should be a limit $$\Phi(B,f)=\lim_{\ldots}\sum_{k=1}^N f(\xi_k){\rm vol}(B_k)\ ,\tag{1}$$ whereby $B=\bigcup_{k=1}^N B_k$ is a partition of $B$ into tiny "almost disjoint" subdomains $B_k$, and for each $k\in[N]$ a sampling point $\xi_k\in B_k$ is chosen. (A lot of technical work is needed to make the $\ldots$ in $(1)$ precise.) In any case the integral is a limit of general Riemann sums defined without ado in a real, complex, or vectorial way.
When it comes to (ii), things are simple: While integrating you may of course describe the points $x\in B$ using your preferred coordinates, e.g., spherical coordinates $r$, $\phi$, $\theta$, and don't forget the Jacobian factor! It is another thing with the values of the function $f$ to be integrated. The linearity of the integral makes it obvious that we can compute the integral of a vector-valued function $f=(f_1,f_2,f_3)$ coordinate-wise. Here it is of course necessary to express the values of $f$ in ordinary cartesian coordinates. It is just not true that $$\left|\int_B f(x)\>{\rm d}(x)\right|=\int_B \bigl|f(x)\bigr|\>{\rm d}(x)\ ,$$ or similar. If, e.g., a complex-valued $f$ is given in the form $f(u,v,w)=r(u,v,w)e^{i\phi(u,v,w)}$ you have to deal with $${\rm Re}f(u,v,w)=r(u,v,w)\cos\phi(u,v,w),\quad {\rm Im}f(u,v,w)=r(u,v,w)\sin\phi(u,v,w)$$ separately.
I think there's a bit of an XY problem here: it seems like your actual problem is to find the gravitational field $\mathbf{g}$ of an irregular object, i.e., to find a $\phi$ satisfying the Poisson equation \begin{align}\nabla^2\phi&=-4\pi G\rho&\phi&=O(\tfrac{1}{r}) \end{align} where $\rho$ is a compactly supported mass distribution—then $\mathbf{g}$ is given by $$\mathbf{g}=-\nabla\phi\text{.}$$ The solution to this equation is $$\phi(\mathbf{x})=-Gm\int_{\mathbb{R}^3}\frac{\rho(\mathbf{x'})\mathrm{d}^3x'}{\lVert\mathbf{x}-\mathbf{x'}\rVert}$$ which is a scalar-valued integral; once this integral has been evaluated, $\mathbf{g}$ can be found by taking the gradient.