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If you have a two-variable function described using polar coordinates, how do you compute its double integral?

## Background

- Polar coordinates (video)
- Double integrals beyond volume

## What we're building to

When you are performing a double integral,

$\iint}_{{R}}f{\textstyle \phantom{\rule{0ex}{0ex}}}{dA$ if you wish to express the function

and the bounds for the region$f$ in polar coordinates${R}$ , the way to expand the tiny area$(r,\theta )$ is$dA$ ${dA}=r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr$ (Pay attention to the fact that the variable

is part of this expression)$r$ Beyond that one rule, these double integrals are mostly about being careful to make sure the bounds of your integrals appropriately encode the region

.$R$ Integrating using polar coordinates is handy whenever your function or your region have some kind of rotational symmetry. For example, polar coordinates are well-suited for integration in a disk, or for functions including the expression

.${x}^{2}+{y}^{2}$

## Example 1: Tiny areas in polar coordinates

Suppose we have a multivariable function defined using the polar coordinates

$f(r,\theta )={r}^{2}$

And let's say you want to find the double integral of this function in the region where

$r\le 2$

This is a disc of radius

Written abstractly, here's what this double integral might look like:

$\begin{array}{r}{\iint}_{r\le 2}{r}^{2}{\textstyle \phantom{\rule{0ex}{0ex}}}{dA}\end{array}$

You could interpret this as the volume underneath a paraboloid (the three-dimensional analog of a parabola), as pictured below:

The question is, what do we do with that

Warning!: You might be tempted to replace with ${dA}$ , since in cartesian coordinates we replace it with $d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr$ . But this is not correct! $dx{\textstyle \phantom{\rule{0ex}{0ex}}}dy$

Remember what a double integral is doing: It chops up the region that we are integrating over into tiny pieces, and

Why did I choose to chop it in this spiderweb pattern, as opposed to using vertical and horizontal lines? Since we are in polar coordinates, it will be easiest to think about the tiny pieces if their edges represent either a constant

Let's focus on just one of these little chunks:

Even though this little piece has a curve shape, if we make finer and finer cuts, we can basically treat it as a rectangle. The length of one side of this "rectangle" can be thought of as

Using a differential

But how long is the other side?

It's not **radians are not a unit of length**. To turn radians into a bit of arc length, we must **multiply by **.

Therefore, if we treat this tiny chunk as a rectangle, and as *is* a rectangle, its area is

${dA}=(r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta )(dr)$

Plugging this into our original integral, we get

$\begin{array}{r}{\iint}_{r\le 2}{r}^{2}{\textstyle \phantom{\rule{0ex}{0ex}}}{dA}={\iint}_{r\le 2}{r}^{2}{\textstyle \phantom{\rule{0ex}{0ex}}}(r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta )(dr)={\iint}_{r\le 2}{r}^{3}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr\end{array}$

Putting bounds on this region is relatively straight-forward in this example, because circles are naturally suited for polar coordinates. Since we wrote

**Concept check**: Evaluate this double integral

$\begin{array}{r}{\int}_{0}^{2}{\int}_{0}^{2\pi}{r}^{3}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr=\end{array}$

$\begin{array}{rl}& {\textstyle \phantom{\rule{0ex}{0ex}}}{\int}_{0}^{2}\underset{\begin{array}{c}{r}^{3}\text{can be factored}\\ \text{out of the inner integral}\\ \text{with respect to}\theta \end{array}}{\underset{\u23df}{\left({\int}_{0}^{2\pi}{r}^{3}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \right)}}dr\\ \\ & ={\int}_{0}^{2}{r}^{3}\left({\int}_{0}^{2\pi}d\theta \right)dr\\ \\ & ={\int}_{0}^{2}{r}^{3}(2\pi )dr\\ \\ & =2\pi {\int}_{0}^{2}{r}^{3}dr\\ \\ & =2\pi {\left({\displaystyle \frac{{r}^{4}}{4}}\right)}_{0}^{2}\\ \\ & =2\pi ({\displaystyle \frac{{2}^{4}}{4}}-{\displaystyle \frac{{0}^{4}}{4}})\\ \\ & =8\pi \end{array}$

## Example 2: Integrating over a flower

Define a two-variable function

$f(r,\theta )=r\mathrm{sin}(\theta )$

Let

$r\le \mathrm{cos}(2\theta )$

Solve the double integral

$\begin{array}{r}{\iint}_{R}f{\textstyle \phantom{\rule{0ex}{0ex}}}dA\end{array}$

**Step 1**: Which of the following represents the right way to replace

The second choice is correct:

$\begin{array}{r}{\iint}_{R}{r}^{2}\mathrm{sin}(\theta ){\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \end{array}$ The function

is defined as $f$ , and the term $r\mathrm{sin}(\theta )$ is expanded as $dA$ (don't forget to add the $r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr$ term here) $r$

$\begin{array}{rl}& {\textstyle \phantom{\rule{0ex}{0ex}}}{\iint}_{R}f{\textstyle \phantom{\rule{0ex}{0ex}}}dA\\ \\ & ={\iint}_{R}\underset{f(r,\theta )}{\underset{\u23df}{r\mathrm{sin}(\theta )}}{\textstyle \phantom{\rule{0ex}{0ex}}}\underset{dA}{\underset{\u23df}{r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr}}\\ \\ & ={\iint}_{R}{r}^{2}\mathrm{sin}(\theta ){\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \end{array}$

**Step 2**: Now we must encode the fact that

The first answer choice is correct

$\begin{array}{r}{\int}_{0}^{2\pi}{\int}_{0}^{\mathrm{cos}(2\theta )}{r}^{2}\mathrm{sin}(\theta ){\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \end{array}$ The inner integral reflects

bounds, since $r$ is written before $dr$ . Our region is explicitly defined such that $d\theta $ , and since these are polar coordinates, $r\le \mathrm{cos}(2\theta )$ can only ever be positive. Therefore, $r$ ranges from $r$ to $0$ . $\mathrm{cos}(2\theta )$ For the outer integral,

ranges over its full range from $\theta $ to $0$ , since no constraints are given on $2\pi $ in the definition of $\theta $ . $R$

**Step 3**: Solve this integral.

$\begin{array}{r}{\int}_{0}^{2\pi}{\int}_{0}^{\mathrm{cos}(2\theta )}{r}^{2}\mathrm{sin}(\theta ){\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta =\end{array}$ In practice, I would use a calculator or Wolfram Alpha to solve one of these integrals. The hard part, which typically requires more human intelligence, is getting the double integral to the point where all the terms are in place. If you want to solve it by hand, here's how it might go:

$\begin{array}{rl}& {\textstyle \phantom{\rule{0ex}{0ex}}}{\int}_{0}^{2\pi}\underset{\text{Factor out}\mathrm{sin}(\theta )\text{from}r\text{-integral}}{\underset{\u23df}{{\int}_{0}^{\mathrm{cos}(2\theta )}{r}^{2}\mathrm{sin}(\theta ){\textstyle \phantom{\rule{0ex}{0ex}}}dr}}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \\ \\ & ={\int}_{0}^{2\pi}\mathrm{sin}(\theta ){\int}_{0}^{\mathrm{cos}(2\theta )}{r}^{2}{\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \\ \\ & ={\int}_{0}^{2\pi}\mathrm{sin}(\theta ){\left[{\displaystyle \frac{{r}^{3}}{3}}\right]}_{0}^{\mathrm{cos}(2\theta )}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \\ \\ & ={\int}_{0}^{2\pi}\mathrm{sin}(\theta )\left({\displaystyle \frac{\mathrm{cos}(2\theta {)}^{3}}{3}}\right){\textstyle \phantom{\rule{0ex}{0ex}}}d\theta \\ \\ & ={\displaystyle \frac{1}{3}}{\int}_{0}^{2\pi}\mathrm{sin}(\theta )\mathrm{cos}(2\theta {)}^{3}d\theta \end{array}$ From here, we can save a lot of computational trouble with an observation: The inside of the integral is an odd function, meaning that when you replace

by $\theta $ , the value of the expression as a whole becomes negative. $-\theta $ Also, since these trigonometric functions are periodic, integrating between

and $0$ is the same as integrating between $2\pi $ and $-\pi $ . The integral on the negative half cancels out with the integral on the positive portion, so in total it is $\pi $ . $0$

## Example 3: The bell curve

Are you ready for one of my favorite results in math? This is really quite clever.

Question: What is the integral ? $\begin{array}{r}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{x}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dx\end{array}$

This single integral is hard-to-impossible to compute directly. Just try to find the antiderivative!

This integral is asking about the area under a bell curve, which turns out to be super important for probability and statistics!

"What does this have to do with double integrals in polar coordinates?"

I hear you, my inquisitive friend, it does seem unrelated, doesn't it? Well, this is where someone got super clever.

Surprisingly, it is easier to solve this multi-dimensional analog of this problem. Namely, find the *volume* under a three-dimensional bell curve over the entire

$\begin{array}{r}{\iint}_{xy\text{-plane}}{e}^{-({x}^{2}+{y}^{2})}{\textstyle \phantom{\rule{0ex}{0ex}}}dA\end{array}$

If we keep everything in cartesian coordinates, this is as hard to solve as the original single integral.

$\begin{array}{r}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-({x}^{2}+{y}^{2})}{\textstyle \phantom{\rule{0ex}{0ex}}}dx{\textstyle \phantom{\rule{0ex}{0ex}}}dy\end{array}$

However, something magical happens when we convert to polar coordinates.

**Concept check**: Express this double integral using polar coordinates.

The second answer choice is correct.

First, make the following two conversions

${x}^{2}+{y}^{2}\to {r}^{2}$ and

$\begin{array}{r}dy{\textstyle \phantom{\rule{0ex}{0ex}}}dx\to \underset{\text{Polar form of}dA}{\underset{\u23df}{r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr}}\end{array}$ So our integral looks like this:

$\begin{array}{r}{\iint}_{r\theta \text{-plane}}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr\end{array}$ To put bounds on the integral, we cover the entire plane by letting

range from $\theta $ to $0$ , and $2\pi $ range from $r$ to $0$ . $\mathrm{\infty}$

$\begin{array}{r}{\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{2\pi}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr\end{array}$

Since the inner integral is with respect to

$\begin{array}{rl}& {\textstyle \phantom{\rule{0ex}{0ex}}}{\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{2\pi}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta {\textstyle \phantom{\rule{0ex}{0ex}}}dr\\ \\ & ={\int}_{0}^{\mathrm{\infty}}{e}^{-{r}^{2}}r\underset{\text{This evaluates to}2\pi}{\underset{\u23df}{{\int}_{0}^{2\pi}{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta}}{\textstyle \phantom{\rule{0ex}{0ex}}}dr\\ \\ & ={\int}_{0}^{\mathrm{\infty}}\left({e}^{-{r}^{2}}r\right)(2\pi ){\textstyle \phantom{\rule{0ex}{0ex}}}dr\\ \\ & =2\pi {\int}_{0}^{\mathrm{\infty}}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr\end{array}$

**Concept check**: Find the antiderivative of

$\begin{array}{r}\int {e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr=\end{array}$

Using u-substitution:Define

as $u$

$u=-{r}^{2}$ which means

$du=-2rdr$ Our indefinite integral now becomes

$\begin{array}{rl}\int {e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr& =\int {e}^{u}r{\textstyle \phantom{\rule{0ex}{0ex}}}{\displaystyle \frac{du}{-2r}}\\ \\ & =\int {e}^{u}{\textstyle \phantom{\rule{0ex}{0ex}}}{\displaystyle \frac{-1}{2}}du\\ \\ & =-{\displaystyle \frac{1}{2}}{e}^{u}\end{array}$ Substituting back to

, this means our final antiderivative is $u=-{r}^{2}$

$\begin{array}{rl}\int {e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr& =-{\displaystyle \frac{1}{2}}{e}^{-{r}^{2}}\end{array}$

Notice, the reason you can now find an antiderivative is because of that little

**Concept check**: Using this antiderivative, finish solving the integral which computes volume under a three-dimensional bell curve.

$\begin{array}{r}2\pi {\int}_{0}^{\mathrm{\infty}}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr=\end{array}$

$\begin{array}{rl}2\pi {\int}_{0}^{\mathrm{\infty}}{e}^{-{r}^{2}}r{\textstyle \phantom{\rule{0ex}{0ex}}}dr& =2\pi {[-{\displaystyle \frac{1}{2}}{e}^{-{r}^{2}}]}_{0}^{\mathrm{\infty}}\\ \\ & =2\pi [-{\displaystyle \frac{1}{2}}{e}^{-(\mathrm{\infty}{)}^{2}}-(-{\displaystyle \frac{1}{2}}{e}^{-{0}^{2}})]\\ \\ & =2\pi [-{\displaystyle \frac{1}{2}}(0)-(-{\displaystyle \frac{1}{2}}(1))]\\ \\ & =2\pi {\displaystyle \frac{1}{2}}\\ \\ & =\pi \end{array}$

Isn't that a beautiful answer? It gets better, you can use this multi-dimensional result to solve the original single integral. Can you see how?

Let

Following the hint, start by saying

$\begin{array}{r}C={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{x}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dx\end{array}$

Now, as we start chewing on the double integral in cartesian coordinates, even though it cannot be solved directly the way it was in polar form, we can start writing things in terms of

$\begin{array}{rl}& {\textstyle \phantom{\rule{0ex}{0ex}}}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-({x}^{2}+{y}^{2})}{\textstyle \phantom{\rule{0ex}{0ex}}}dx{\textstyle \phantom{\rule{0ex}{0ex}}}dy\\ \\ & ={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{x}^{2}}{e}^{-{y}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dx{\textstyle \phantom{\rule{0ex}{0ex}}}dy\\ \\ & ={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{y}^{2}}\underset{\text{This equals}C}{\underset{\u23df}{{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{x}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dx}}{\textstyle \phantom{\rule{0ex}{0ex}}}dy\\ \\ & ={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{y}^{2}}C{\textstyle \phantom{\rule{0ex}{0ex}}}dy\\ \\ & =C\underset{\text{This also equals}C}{\underset{\u23df}{{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{y}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dy}}\\ \\ & ={C}^{2}\end{array}$

In other words, even if we don't know what the area under a bell curve is, we know that when you square it, you get the volume under a three-dimensional bell curve. But we just solved the volume under three-dimensional bell curve using polar-coordinate integration! We found that the volume was

$\begin{array}{r}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-{x}^{2}}{\textstyle \phantom{\rule{0ex}{0ex}}}dx=\sqrt{\pi}\end{array}$

Isn't that crazy? This expression looks like it has nothing to do with

## Summary

The only real thing to remember about double integral in polar coordinates is that

$dA=r{\textstyle \phantom{\rule{0ex}{0ex}}}dr{\textstyle \phantom{\rule{0ex}{0ex}}}d\theta $ Beyond that, the tricky part is wrestling with bounds, and the nastiness of actually solving the integrals that you get. But those are the same difficulties one runs into with cartesian double integrals.

The reason this is worth learning is that sometimes double integrals become simpler when you phrase them with polar coordinates, as was the case in the bell curve example.

## Want to join the conversation?

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maycol.medina

6 years agoPosted 6 years ago. Direct link to maycol.medina's post “I read in Wikipedia that ...”

I read in Wikipedia that in order to solve the guassian integral you have to make use of Fubini's Theorem, so can somene please give me a simple explanation of what that theorem is about?

•

(3 votes)

Coolin

5 years agoPosted 5 years ago. Direct link to Coolin's post “This is (10 months) late,...”

This is (10 months) late, but I figured I'd still answer for anyone wandering about the same question.

There are quite a few different ways to solve the Gaussian integral. The "standard" way does not need to use Fubini's theorem, however there are several other ways that do.

Fubini's theorem deals with when you can interchange integrals. In short, if you replace the integrand with its absolute value, and you obtain a finite value when you perform the double integral, then you can freely interchange the order of integrations.(4 votes)

bhattacharyajhumki

8 years agoPosted 8 years ago. Direct link to bhattacharyajhumki's post “how can we take limits in...”

how can we take limits in double integral through polar coordinates without drawing the diagram?

•

(4 votes)

jbmxtmx

4 years agoPosted 4 years ago. Direct link to jbmxtmx's post “I strongly recommend you ...”

I strongly recommend you to draw a diagram of the integration region. That way you reduce the likelihood of making a mistake in your limits of integration.

(1 vote)

Ritish Maram

7 years agoPosted 7 years ago. Direct link to Ritish Maram's post “n other words, even if we...”

n other words, even if we don't know what the area under a bell curve is, we know that when you square it, you get the volume under a three-dimensional bell curve. But we just solved the volume under three-dimensional bell curve using polar-coordinate integration! We found that the volume was

π

πpi. Therefore, the original integral is

π

√

π

square root of, pi, end square root.

MAY I HAVE DETAILED EXPLANATION OF THE ABOVE MENTIONED PARAGRAPH?•

(3 votes)

Dylan Yu

6 years agoPosted 6 years ago. Direct link to Dylan Yu's post “Let the area of the bell ...”

Let the area of the bell curve be C. Then C^2 is a double integral that is easy to solve in polar coordinates. After computing C^2, we take the square root to find C, the area of the bell curve.

(2 votes)

NEEMISH TEJASWI

3 years agoPosted 3 years ago. Direct link to NEEMISH TEJASWI's post “In example 2, we first in...”

In example 2, we first integrate over r and then over theta. How would the integration limits change if we decide to integrate over theta first followed by integration over r? I understand that in reality evaluating this integral might be very difficult/ impractical but nevertheless, I would at least like to formulate the problem.

•

(3 votes)

Richard

5 years agoPosted 5 years ago. Direct link to Richard's post “I am having some trouble ...”

I am having some trouble with the second example. If you accept the author's statement "since these are polar coordinates, r can only ever be positive" then if you graph r=cos(2*theta) you only get two lobes. If I accept r going negative, then do I obtain "signed" volumes as in single variable calc I can obtain signed areas? That way I can see the suggested answer of zero.

•

(3 votes)

Hexuan Sun 9th grade

a year agoPosted a year ago. Direct link to Hexuan Sun 9th grade's post “How do you graph polar eq...”

How do you graph polar equations in 3D? Because there's 2 inputs now.

•

(1 vote)

Venkata

a year agoPosted a year ago. Direct link to Venkata's post “Correct. You cannot use p...”

Correct. You cannot use polar coordinates to express a point in 3D.

To fix this, we add a third coordinate: z. Yeah, it's the same "z" you're used to- height above the xy plane. So, for 3D, we use the coordinates (r,θ,z). However, we don't call this coordinate system polar anymore. It's called the "cylindrical coordinate system", and you'll use it to integrate, well, cylinders with triple integrals. You'll also see a new coordinate system called the "spherical coordinate system" which is used for spheres and even cones

(4 votes)

Tiago Sutter

7 years agoPosted 7 years ago. Direct link to Tiago Sutter's post “There is a link in the te...”

There is a link in the text at "Background" (Polar coordinates (video)) section, but the page doesn't exist

•

(2 votes)

gschex1112

7 years agoPosted 7 years ago. Direct link to gschex1112's post “Yeah, the closest thing I...”

Yeah, the closest thing I found was this, but I don't think that's the right one.

https://www.khanacademy.org/math/calculus-home/integration-applications-calc/area-defined-by-polar-graphs-calc/v/formula-area-polar-graph(2 votes)

Müberra Bodur

3 years agoPosted 3 years ago. Direct link to Müberra Bodur's post “why do we multiply with 2...”

why do we multiply with 2pi? is it always a step in polar coordinates?

•

(1 vote)

adam.ghatta

2 years agoPosted 2 years ago. Direct link to adam.ghatta's post “The 2pi in the upper boun...”

The 2pi in the upper bound comes from the fact that this polar curve has a domain for theta being in [0,2pi], since it makes one revolution once it reaches 2pi.

(2 votes)

White

7 years agoPosted 7 years ago. Direct link to White's post “Is there a formulation fo...”

Is there a formulation for this using the area of a sector?

•

(1 vote)

bh3sk3r

6 years agoPosted 6 years ago. Direct link to bh3sk3r's post “If we use area of a secto...”

If we use area of a sector, we no longer have to integrate over r, only over theta. The solution is no longer a double integral. The incremental sector in this case is approximately a triangle and dA = (1/2)(r)(rdθ). We can verify this gives the area of a circle correctly by replacing r with its radius (a constant) and setting the limit of integration from 0 to 2π .

(2 votes)

kMoney

a year agoPosted a year ago. Direct link to kMoney's post “I don't understand how th...”

I don't understand how the radial distance becomes rdθ. Arc length = diameter*(θ/2pi) = 2r(θ/2pi) = rθ/pi. Why does the pi disappear?

•

(1 vote)

Venkata

a year agoPosted a year ago. Direct link to Venkata's post “Arc length equals $(\thet...”

Arc length equals $(\theta)/2(\pi) \cdot (\pi)(diameter)$. You missed the second $\pi$. So, the $\pi$ terms cancel out and you get r$\theta$.

(2 votes)