# Integration in Polar Coordinates (Fubini/Tonelli)

• MHB
• joypav
In summary: So in the first integral, it is $\nu$ and in the second one it is $\lambda$. Sorry for the confusion.
joypav
Let $S^{n-1} = \left\{ x \in R^2 : \left| x \right| = 1 \right\}$ and for any Borel set $E \in S^{n-1}$ set $E* = \left\{ r \theta : 0 < r < 1, \theta \in E \right\}$. Define the measure $\sigma$ on $S^{n-1}$ by $\sigma(E) = n \left| E* \right|$.

With this definition the surface area $\omega_{n-1}$ of the sphere in $R^n$ satisfies $\omega_{n-1} = n \gamma_n = \frac{2 \pi^{n/2}}{\Gamma (n/2)}$, where $\gamma_n$ is the volume of the unit ball in $R^n$. Prove that for all non-negative Borel functions $f$ on $R^n$,

$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr$.

In particular, if $f$ is a radial function, then ($f$ radial $\implies f(x) = f(\left| x \right|)$)

$\int_{R^n} f(x) dx = \frac{2 \pi^{n/2}}{\Gamma (n/2)} \int_0^{\infty} r^{n-1} f(r) dr = n \gamma_n \int_0^{\infty} r^{n-1} f(r) dr$.

I admit my head is spinning over this problem. Can someone give me an idea of the proof or explain the idea of it more clearly...
Also, this problem is in the section for Fubini/Tonelli Theorems.

joypav said:
Let $S^{n-1} = \left\{ x \in R^2 : \left| x \right| = 1 \right\}$ and for any Borel set $E \in S^{n-1}$ set $E* = \left\{ r \theta : 0 < r < 1, \theta \in E \right\}$. Define the measure $\sigma$ on $S^{n-1}$ by $\sigma(E) = n \left| E* \right|$.

With this definition the surface area $\omega_{n-1}$ of the sphere in $R^n$ satisfies $\omega_{n-1} = n \gamma_n = \frac{2 \pi^{n/2}}{\Gamma (n/2)}$, where $\gamma_n$ is the volume of the unit ball in $R^n$. Prove that for all non-negative Borel functions $f$ on $R^n$,

$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr$.

In particular, if $f$ is a radial function, then ($f$ radial $\implies f(x) = f(\left| x \right|)$)

$\int_{R^n} f(x) dx = \frac{2 \pi^{n/2}}{\Gamma (n/2)} \int_0^{\infty} r^{n-1} f(r) dr = n \gamma_n \int_0^{\infty} r^{n-1} f(r) dr$.

I admit my head is spinning over this problem. Can someone give me an idea of the proof or explain the idea of it more clearly...
Also, this problem is in the section for Fubini/Tonelli Theorems.
$\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\vp}{\varphi}$
$\newcommand{\mc}{\mathcal}$
$\newcommand{\R}{\mathbf R}$Let $(X, \mc X, \mu)$ be a measure space and $(Y, \mc Y)$ be a measurable space and let $\vp:X\to Y$ be a measurable function.
We define the push-forward of $\mu$ via $\vp$ as a measure on $(Y, \mc Y)$, which we denote by $\vp_*\mu$, by declaring $(\vp_*\mu)(E)=\mu(\vp^{-1}(E))$ for all $E\in \mc Y$.

A key theorem, which is easy to prove, is that if $g:Y\to \R$ is a measurable function such that $g\circ f$ is an integrable function on $X$, the we have

\int_X g\circ \vp\ d\mu = \int_Y g\ d(\vp_*\mu)

We will use this theorem to solve the problem at hand.
We have a natural map $\vp: (0, \infty)\times S^{n-1}\to \R^n\setminus\set{0}$ which sends $(r, \theta)$ to $r\theta$.
Write $\nu$ to denote the Lebesgue measure on $\R^n\setminus \set{0}$.
Define a measure $\mu$ on $(0, \infty)\times S^{n-1}$ as $\mu=\vp^{-1}_*\nu$.
The main observation is that if $\lambda$ is the Lebesgue measure on $\R$, then the measure $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.

So we have

\int_{\R^n}f\ d\nu =\int_{\R^n\setminus \set{0}} f\ d\nu = \int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma)

Now by Fubini

\int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma) = \int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)

which is, by the fact that $d\alpha=x^{n-1} \ d\lambda$, is

\int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)=\int_0^\infty r^{n-1} \left(\int_{S^{n-1}} f(r\theta) \ d\sigma(\theta)\right) d\lambda(r)

Thank you!
Makes sense. I wonder why we haven't seen the "push forward"? (Wondering)

In the very first integral, is that meant to be $d\lambda$?

joypav said:
Thank you!
Makes sense. I wonder why we haven't seen the "push forward"? (Wondering)

In the very first integral, is that meant to be $d\lambda$?
I am using $\nu$ to denote the Lebesgue measure on $\mathbb R^n$ and $\lambda$ to denote the Lebesgue measure on $\mathbb R$.

caffeinemachine said:
$\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\vp}{\varphi}$
$\newcommand{\mc}{\mathcal}$
$\newcommand{\R}{\mathbf R}$Let $(X, \mc X, \mu)$ be a measure space and $(Y, \mc Y)$ be a measurable space and let $\vp:X\to Y$ be a measurable function.
We define the push-forward of $\mu$ via $\vp$ as a measure on $(Y, \mc Y)$, which we denote by $\vp_*\mu$, by declaring $(\vp_*\mu)(E)=\mu(\vp^{-1}(E))$ for all $E\in \mc Y$.

A key theorem, which is easy to prove, is that if $g:Y\to \R$ is a measurable function such that $g\circ f$ is an integrable function on $X$, the we have

\int_X g\circ \vp\ d\mu = \int_Y g\ d(\vp_*\mu)

We will use this theorem to solve the problem at hand.
We have a natural map $\vp: (0, \infty)\times S^{n-1}\to \R^n\setminus\set{0}$ which sends $(r, \theta)$ to $r\theta$.
Write $\nu$ to denote the Lebesgue measure on $\R^n\setminus \set{0}$.
Define a measure $\mu$ on $(0, \infty)\times S^{n-1}$ as $\mu=\vp^{-1}_*\nu$.
The main observation is that if $\lambda$ is the Lebesgue measure on $\R$, then the measure $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.

So we have

\int_{\R^n}f\ d\nu =\int_{\R^n\setminus \set{0}} f\ d\nu = \int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma)

Now by Fubini

\int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma) = \int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)

which is, by the fact that $d\alpha=x^{n-1} \ d\lambda$, is

\int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)=\int_0^\infty r^{n-1} \left(\int_{S^{n-1}} f(r\theta) \ d\sigma(\theta)\right) d\lambda(r)
sorry if I am re opening this , I have the same problem , can you explain why $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.? Really thanks

## 1. What is integration in polar coordinates?

Integration in polar coordinates is a method of finding the area under a curve in a polar coordinate system. It involves converting the given function into polar coordinates and then using the Fubini or Tonelli theorem to evaluate the integral.

## 2. What is the Fubini theorem?

The Fubini theorem states that if a function is continuous over a region in a coordinate system, then the integral of that function over that region can be calculated by integrating first with respect to one variable and then with respect to the other variable.

## 3. What is the Tonelli theorem?

The Tonelli theorem is a generalization of the Fubini theorem that allows for the integration of non-negative functions over a product measure space. It states that if a function is measurable over a product measure space, then the integral of that function can be calculated by integrating first with respect to one variable and then with respect to the other variable.

## 4. What are the advantages of using polar coordinates for integration?

Using polar coordinates for integration can make it easier to evaluate integrals with circular or symmetric boundaries. It can also simplify the calculation of integrals involving trigonometric functions.

## 5. What are some practical applications of integration in polar coordinates?

Integration in polar coordinates has many practical applications in fields such as physics, engineering, and mathematics. It is commonly used to calculate the area of irregular shapes, the volume of 3D objects, and the center of mass of a system. It is also used in the study of electric and magnetic fields, fluid mechanics, and signal processing.

• Topology and Analysis
Replies
4
Views
507
• Topology and Analysis
Replies
38
Views
3K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
4
Views
484
• Topology and Analysis
Replies
1
Views
409
• Topology and Analysis
Replies
2
Views
299
• Topology and Analysis
Replies
2
Views
542
• Topology and Analysis
Replies
4
Views
2K
• Topology and Analysis
Replies
4
Views
543
• Topology and Analysis
Replies
2
Views
809