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Whenever I have been presented with integrals as (*), I have always used some sort of symmetry to get around actually calculating the integral. Now it just struck me that I have no idea how to "mechanically" calculate (*):

$$ \iiint \left(f(r)\hat{r}\right)\left(r^2\sin\phi\right) drd\phi d\theta \qquad \textbf{(*)} $$ If it is necessary to define what region you are talking about for the question to make sense, I'll pick the unit sphere.

Would this have been the integral $$ \int f(x)\hat{x} dx $$ I would have proceeded by moving the unit vector outside of the integral and calculated as usual

$$ \hat{x}\int f(x)\cdot dx = \hat{x}\left(F(x) +C\right) $$

However, the problem with the integral is that the direction of $\hat{r}$ changes as $\phi$ and $\theta$ changes. Therefore I don't see how I could treat the unit vector as a constant and have no idea how to start with (*).

I apologize in advance if this turns out to be a duplicate. How to integrate a vector function in spherical coordinates? verifies my gut feeling that $\hat{r}$ is not constant but doesn't make things much clearer for me than that.

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  • $\begingroup$ But the unit vector $\hat x$ likewise changes with $x$? Why can you take it outside the integral? $\endgroup$
    – joriki
    Aug 24, 2015 at 16:25
  • $\begingroup$ I am having a hard time understanding how that could be. In a Cartesian coordinate system, I would have thought that $\hat{x}$ never changes. It points in the same direction all the time. $\endgroup$
    – V.Vocor
    Aug 24, 2015 at 16:34
  • $\begingroup$ Then perhaps you're using some other convention. Under the convention that I was applying, $\hat x$ is the unit vector that points along the position vector $x$. Clearly the position vector $x$ doesn't always point in the same direction. $\endgroup$
    – joriki
    Aug 24, 2015 at 16:39
  • $\begingroup$ I meant for $\hat{x}$ to be the direction of the x-axis in a fixed coordinate system. If there's is a less confusing way for me to write that, please, let me know. $\endgroup$
    – V.Vocor
    Aug 24, 2015 at 16:45
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    $\begingroup$ Makes sense. My way is unnecessarily ambiguous. I'll try to adopt your approach in the future. Thank you, again. $\endgroup$
    – V.Vocor
    Aug 24, 2015 at 17:07

1 Answer 1

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Since $\hat{\mathbf{r}} = \sin\theta\cos\phi\,\hat{\mathbf{x}} + \sin\theta\sin\phi\,\hat{\mathbf{y}} + \cos\theta\,\hat{\mathbf{z}}$, you can write

$$ \iiint f(r)\,\hat{\mathbf{r}}\,r^2\sin\theta\,dr\,d\phi\,d\theta = \iiint f(r) \left(\sin\theta\cos\phi\,\hat{\mathbf{x}} + \sin\theta\sin\phi\,\hat{\mathbf{y}} + \cos\theta\,\hat{\mathbf{z}}\right)r^2\sin\theta\,dr\,d\phi\,d\theta $$ which is the sum of three separate integrals: $$ \hat{\mathbf{x}} \iiint f(r)\sin\theta\cos\phi\,r^2\sin\theta\,dr\,d\phi\,d\theta = \hat{\mathbf{x}} \int f(r)\,r^2\,dr \int\sin^2\theta\,d\theta \int\cos\phi\,d\phi $$ $$ \hat{\mathbf{y}}\iiint f(r)\,\sin\theta\sin\phi\,r^2\sin\theta\,dr\,d\phi\,d\theta = \hat{\mathbf{y}} \int f(r)\,r^2\,dr \int\sin^2\theta\,d\theta \int\sin\phi\,d\phi $$ $$ \hat{\mathbf{z}}\iiint f(r)\,\cos\theta\,r^2\sin\theta\,dr\,d\phi\,d\theta = \hat{\mathbf{z}} \int f(r)\,r^2\,dr \int\sin\theta\cos\theta\,d\theta \int d\phi $$

where $\hat{\mathbf{r}}$ is the unit vector along the radial direction but $\hat{\mathbf{x}}$, $\hat{\mathbf{y}}$, and $\hat{\mathbf{z}}$ are the Cartesian unit vectors.

(Note that I'm using the convention where the volume element is $r^2\sin\theta\,dr\,d\phi\,d\theta\,$.)

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  • $\begingroup$ Note that there are different conventions for spherical coordinates; the OP's volume element was correct in one of them. $\endgroup$
    – joriki
    Aug 24, 2015 at 16:45
  • $\begingroup$ Fair enough. I updated the answer accordingly. Thanks for pointing that out. $\endgroup$
    – wltrup
    Aug 24, 2015 at 16:49
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    $\begingroup$ Is there a way to calculate the integral without being forced into rectangular coordinates at any point along the way? $\endgroup$ Nov 14, 2017 at 23:42
  • $\begingroup$ @RobinNewhouse I'm afraid not. vectors exist on a tangent space, and only when you have rectangular coordinates do you actually get a "global" vector space via an isomorphism. Otherwise, you can't have position vectors $\endgroup$
    – Max0815
    Jan 24 at 12:31

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