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When finding the average value of a polar function $r(\theta)$ over the interval $[a,b],$ why must we take $$\frac1{b-a}\int_a^b r(\theta) \,\mathrm d\theta\tag1$$ rather than $$\frac1{b-a} \cdot \frac12 \int_a^b r^2(\theta) \,\mathrm d\theta \quad?\tag2$$ Aren't both of them the area of the sector divided by the length of the interval?

ryang
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ada
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  • By dimensional analysis, (1) isn't an area. – J.G. Apr 12 '23 at 07:25
  • @J.G. "The area of a sector of radius $r$ cm and subtending $\theta$ rad at the centre is $\frac12 r^2\theta$ cm$^2.$" Here, as in the above, the variables represent unitless (thus pure, dimensionless) numbers. – ryang Apr 12 '23 at 08:18
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    @ryang Which source are you quoting? Its nondimensionalization of anything costs us the ability to tell certain equations are obvious wrong. The usual convention is to say a radius $r$ is dimensionful, not that a radius is $r$ copies of your favourite unit of length for some dimensionless $r$. – J.G. Apr 12 '23 at 08:20
  • @Dominique Disagree. $\frac{1}{b-a}\int_a^bf(t)dt$ averages $f$, so has the same dimension as $f$, regardless of the dimension of $t$. – J.G. Apr 12 '23 at 09:18
  • @J.G. could you please elaborate on your first comment (that by dimensional analysis, (1) isn't an area)? I think that is the key source of my confusion.

    Because, when in the cartesian plane, the integral of f(x)dx gives the area under the curve, so I was wondering why the same is not true for the integral of r(theta)dtheta in the polar plane.

     – ada Apr 12 '23 at 13:19
  • If you plot $f$ against $x$, $\int_a^bf(x)dx$ is the size of a region in your plot you would describe in said diagram as an area. But the point here of dimensional analysis is this: $I=\int_a^bf(x)dx$ has $a,,b,,x$ all of the same dimension, $I$ of the same dimension as e.g. $af(a)$, and $J=\frac{I}{b-a}$ just of the same dimension as $f(a)$. So if $\theta$ is a dimensionless angle and $r$ is a length (denoted $[r]=\mathsf{L}$), the case $f=r^p$ gives $[J]=\mathsf{L}^p$. "An area" has dimension $\mathsf{L}^2$. – J.G. Apr 12 '23 at 13:35
  • Makes sense, thank you! – ada Apr 12 '23 at 19:17

1 Answers1

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  1. $$\frac1{b-a} \cdot \frac12 \int_a^b r^2(\theta) \,\mathrm d\theta $$

    Since $[f(x)]^2$ and $f^2(x)$ have different meanings (squaring a function versus composing a function with itself), $[r(\theta)]^2$ is best not written as $r^2(\theta).$

    In fact, $r(\theta),$ which is analogous to $y(x),$ is also less clear than $r=f(\theta),$ which is analogous to $y=f(x).$

  2. Since (putting $r=\theta;a=0;b=1$ as an example) $$\frac12=\int_0^1 \theta \,\mathrm d\theta\ne \frac12 \int_0^1 \theta^2\,\mathrm d\theta=\frac16,$$ thus $$\boxed{\int_a^b r \,\mathrm d\theta}\not\equiv \boxed{\frac12 \int_a^b r^2\,\mathrm d\theta};$$ therefore, clearly, the two expressions do not describe the same area.

    Indeed, the LHS expression

    • gives the area bounded by the curve $r=f(\theta)$ and the three lines $\theta=a,\:\theta=b$ and $r=0$ on the $\boldsymbol{(\theta,r)}$ Cartesian plane,

      which is the same area bounded by the curve $y=f(x)$ and the three lines $x=a,\:x=b$ and $y=0$ on the $\boldsymbol{(x,y)}$ Cartesian plane,

    whereas for $p\in\mathbb R$ and $a,b\in[p,p+2\pi),$ the RHS expression

    • gives the area bounded by the curve $r=f(\theta)$ and the two rays $\theta=a$ and $\theta=b$ on the $\boldsymbol{(r,\theta)}$ polar plane,

      which is the same area bounded by the curve $y=\frac12[f(x)]^2$ and the three lines $x=a,\:x=b$ and $y=0$ on the $\boldsymbol{(x,y)}$ Cartesian plane.

    Different coordinate systems (and notice that the blue area is thrice the green area even though both are over same $\theta$-interval):

    enter image description here

  3. Since the rectangular area

    • the average value of $f(\theta)\:$ as $\theta$ varies over $[a,b]\quad\times\quad$ the length of $[a,b]$

    gives the area $\displaystyle\int_a^b f(\theta) \,\mathrm d\theta$ on the Cartesian plane rather than the area $\displaystyle\frac12 \int_a^b [f(\theta)]^2\,\mathrm d\theta$ on the Cartesian plane, the average value of $\boldsymbol r$ over $\boldsymbol{[a,b]}$ is given by $$\frac1{b-a}\int_a^b f(\theta) \,\mathrm d\theta\tag✓$$ rather than $$\displaystyle\frac12\cdot\frac1{b-a}\int_a^b [f(\theta)]^2 \,\mathrm d\theta.\tag✘$$

    To be clear that this visualisation is not using polar coordinates, observe that

    • the average value of $f(x)\:$ as $x$ varies over $[a,b]\quad\times\quad$ the length of $[a,b]$
      $\displaystyle = \int_a^b f(x)\,\mathrm dx$
      $\displaystyle\ne \frac12 \int_a^b [f(x)]^2\,\mathrm dx.$

Addendum

why do you graph the LHS in the cartesian plane while graphing the RHS in the polar plane?

@ada The LHS is not terribly meaningful on the polar plane whereas I'm sure you agree that it represents the usual area under its integrand's Cartesian curve. On the other hand, observe that transplanting the RHS area between the two planes (as described in my second bullet point) changes not its value/size, only its shape.

ryang
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  • Hello, thank you for your reply! Sorry, but I still don't really understand why the LHS and RHS give different values, since aren't they both finding the area bounded by the polar curve? – ada Apr 12 '23 at 02:03
  • No, only the RHS integral gives the polar area. I've expanded the answer, but please allow yourself to disabuse of the wrong notion before reading it. – ryang Apr 12 '23 at 07:20
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    I don't agree that $[r(\theta)]^2$ shouldn't be written as $r^2(\theta)$ for fear of conflation with $r(r(\theta))$. The latter can be denoted $r^{\circ2}(\theta)$. – J.G. Apr 12 '23 at 07:24
  • @J.G. While $\sin^2(x):=[\sin(x)]^2$ is perfectly conventional, for $y^2=[f(x)]^2,$ I prefer not to deal with $f^2(x)$ or $y^2(x)$ or $y(x)^2$ or $f(x)^2$ (regardless of the fact that many folks do like that final option). – ryang Apr 12 '23 at 07:34
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    This goes much further than trigonometry. Since $r$ is a variable equal to a function of $\theta$ rather than just the function itself, $r(r(\theta))$ is an incredibly unrealistic interpretation of $r^2$ (and that's before I get into dimensional analysis and argue such composition is illegal). $r^2$ should be as easy to write down as $\theta^2$. – J.G. Apr 12 '23 at 07:46
  • @J.G. The point is that $x^2,y^2,\theta^2,r^2$ are all fine, whereas $r(\theta)$ and $r^2(\theta)$ are deliberately displaying $r$ as a function rather than a simple variable. Anyhow, this is a minor point because yes, usually disambiguation doesn't take much effort given the context. The point of my Answer's preamble is to be crystal clear about the notation that I am using in point #2 and to start the OP thinking as a setup for the the change-of-variable idea in point #1. – ryang Apr 12 '23 at 08:00
  • So, it comes down to the difference between finding areas of sectors vs rectangles? – ada Apr 12 '23 at 15:48
  • @ada Eh, which expression/concept are you now referring to as a rectangle's area? (Do notice that $16$cm$^2$ is both the area of a square and of a 2cm-by-8cm rectangle; it's all a matter of framing.) $\quad$ P.S. Regarding your new comment way above: no, dimensional analysis is unrelated to—and in fact distracting from—your actual confusion. Understand that $\displaystyle\frac12 \int_a^b r^2,\mathrm d\theta$ has a fixed value for a fixed function $r$ and fixed numbers $a$ and $b,$ but different representations on the polar and Cartesian planes: yes it does look like a sector in the former. – ryang Apr 12 '23 at 16:10
  • I think I mostly get your answer, but why do you graph the LHS in the cartesian plane while graphing the RHS in the polar plane? – ada Apr 12 '23 at 17:04
  • Okay, that makes sense. Thank you for the help! (I did upvote and accept, not sure if the upvote worked since my account is new.) – ada Apr 12 '23 at 19:15