Aren't $\int_a^b r(\theta) \,\mathrm d\theta$ and $\frac12 \int_a^b r^2(\theta) \,\mathrm d\theta$ the same area?

Question

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?

Answer 1

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

   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 $y=f(x).$

1. 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):

   

2. 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 $[a,b]$
     $\displaystyle = \int_a^b f(x)\,\mathrm dx$
     $\displaystyle\ne \frac12 \int_a^b [f(x)]^2\,\mathrm dx.$

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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.