How to convert cartesian domain and ranges into polar

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How do I convert domains and ranges into complex domains? What about domains with infinite restrictions?

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Hello!

[strange belfry]

Hi!

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What do you mean by infinite restrictions?

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You are talking about going from Cartesian to polar?

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Stuff like $(-\infty ,x]$

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Yes

[manic canyon]

cool looking building

[vivid mantle]

That's no restriction. That's an interval involving infinity.

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Oh

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Ohh well we do parameterization for that. You know what it means

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Uh

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?

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No

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Okay so let me take an example.

Take a x, y Cartesian plane

Draw a circle of center O (centered at the origin)

You want to locate every point (x, y) on that circle

For now you’re following?

[strange belfry]

Yep.

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What you would want to do is use the « Cartesian » equation of the circle.

$x^2 + y^2 = R^2$

[manic canyon]

AstroGuy

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You know that equation

?

[strange belfry]

Yep

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but it looks very messy so instead you transfer your Cartesian coordinates (x,y) to polar coordinates

You do it as follows:

$x=Rcos(\theta)$
$y=Rsin(\theta)$

[manic canyon]

AstroGuy

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R represents the radius of the circle

And theta the angle at which the point on your circle is at from the positive x-axis

From the Cartesian equation:

$y=\pm \sqrt{R^2 -x^2}$

[manic canyon]

AstroGuy

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By using Cartesian coordinates, every point has coordinates:

$M=(x, \pm \sqrt{R^2 - x^2})$

[manic canyon]

AstroGuy

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But now transferring to polar you get:

$M=(Rcos(\theta), Rsin(\theta))$

[manic canyon]

AstroGuy

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The new variable here is theta

Not x

x can go from -R to R that’s the domain if you stay on the circle

If x=R then:

R=Rcos(theta)
cos(theta)=1
theta=0, 2pi, 4pi

I think I lost you

[strange belfry]

Nah

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The circle you see on the xy plane is a function y=+-sqrt(R^2-x^2)

If you stay in Cartesian

,w plot sqrt(1-x^2) from x=-1 to x=1

[manic canyon]

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Here

If R=1

You have this

,w plot -sqrt(1-x^2) from x=-1 to x=1

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That’s the bottom part of the circle

You see I used Cartesian

[strange belfry]

Yep

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Only in terms of x

But we know
x=Rcos(theta)

And

y=Rsin(theta)

You can transfer to polar coordinates (theta)

[strange belfry]

Okay

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,w plot sqrt(1-cos(theta)^2) from theta=0 to theta=pi

Okay wolfram Nevermind

So basically:

in Cartesian:

M=(x, +-sqrt(R^2 - x^2)

In polar:

M=(Rcos(theta), Rsin(theta))

You go from x to theta

Different variable

[strange belfry]

Okay

[normal frost]

,w polar plot r = \sqrt( 1-cos(theta)^2) from theta = 0 to pi

[wide widget]

Okay yeah that ahah

Not used to plot parametric with wolfram

[normal frost]

xD

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I know it may have been confusing the way I explained

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[strange belfry]

It's all good, I think I got the gist of it

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But your question was about domain

Remember that
x=Rcos(theta)

If x=R then:

R=Rcos(theta)

1=cos(theta)

theta=2kpi with k an integer

That’s how you get the corresponding domain

If you know the domain of x then you know the domain of theta

[strange belfry]

Yeah

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x=1 => theta=2kpi

You just substitute

[strange belfry]

Okay, how can this apply to unrestricted domain?

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What do you mean unrestricted domain

Once you know the relation between x and theta then you know everything

[strange belfry]

Ah okay

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Imagine

x=2 * theta

x belongs to [0, +infinity[

For now you follow?

[strange belfry]

Ywah

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Replace x by 0

You get 0= 2 * theta

then theta=0

So the lower bound of the domain for theta is also 0

then replace x by +infinity

+infinity = 2 * theta

theta= + infinity / 2 = +infinity

So theta belongs to [0, +infinity[

Same domain

[strange belfry]

Okay

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In that case x and theta have the same domain

[strange belfry]

Okay, cool

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It’s just about knowing: when x= something what is theta equal to?

[strange belfry]

Yeah, that makes sense

About range, do we apply the same logic to y?

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Yup

[strange belfry]

Alright, nice

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And it is very very useful for integrals

You know about integrals

?

[strange belfry]

Yeah, I'm asking this because I'm stumped on an integral

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Ohhh

Okay then you must be wondering what bounds to put on the integrals

[strange belfry]

Uhuh

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You do as I did

If x=3 then theta=…

If x=1 then theta =…

You switch

[strange belfry]

If one of the bounds of a definite integral is x, do I just substitute in 2\theta?

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Yes

[strange belfry]

Yeah, makes sense

Okay, so put x and y into terms of one cartesian vsriable, and just convert to polar is the gist of this, right?

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you mean you have x and y. And you wonder if you should convert them to polar?

[strange belfry]

I'm wondering if the process of converting a domain/range into polar is just putting both cartesian coordinates into terms of one variable, snd then converting to polar.

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Well if you put them both into terms of one variable then you already transferred your coordinates to another coordinate system

In this case polar system

your new variable becomes theta

[strange belfry]

Oh, I mean converting x into terms of y or y into terms of x

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Ohhhh

Sorry

[strange belfry]

No, u good

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For some integrals, yes it is better to express y(x)

Then convert x into polar

$\int_0^{R} ydx$

[manic canyon]

AstroGuy

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That will give you area between the curve y and the x axis

From x=0 to x=R

Now it’s time to write y(x)

$\int_0^{R} \sqrt{R^2-x^2}dx$

[manic canyon]

AstroGuy

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You understand for now?

[strange belfry]

Yeah

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Im still in Cartesian so I don’t touch to the bounds yet

Now it’s time to switch to polar using the parameterization:

x=Rcos(theta)

$dx=-Rsin(\theta)d\theta$

You understand this part?

[manic canyon]

AstroGuy

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Okay, so how do we make that conversion of the differential?

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You express dx in terms of dtheta

Then you plug in the integral

[strange belfry]

Yes, I'm aware, but why not just do $R\sin(d\theta?)$

[manic canyon]

cool looking building

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Because that’s not what dx is equal to

Look:

$x=Rcos(\theta)$

[manic canyon]

AstroGuy

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Then:

$\frac{dx}{d\theta}=-Rsin(\theta)$

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AstroGuy

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Derivative of x with respect to theta

You agree with this?

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Yeah I agree

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Then:

$dx=-Rsin(\theta)d\theta$

[manic canyon]

AstroGuy

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Ohhh

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You see?

[strange belfry]

Yeah, I see now

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You can start from the derivative

Then multiply by d(theta) to get dx

Now the bounds

$\int_0^{R} \sqrt{R^2-x^2}dx$

[manic canyon]

AstroGuy

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x=0 to x=R you agree?

[strange belfry]

Yep

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So use x=Rcos(theta)

If x=0:

0=Rcos(theta)

Then theta=pi/2

You agree?

[strange belfry]

Okay

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If x=R then:

R=Rcos(theta)
1=cos(theta)
theta=0

[strange belfry]

Yeah

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Okay so now we have everything

You plug everything

It becomes:

$\int_\frac{\pi}{2}^{0} \sqrt{R^2 - R^2cos(\theta)^2}(-Rsin(\theta))d\theta$

Oh wait

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AstroGuy

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Do you agree with that

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Yeah

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You sure?

[strange belfry]

100%

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Okay okay

We can rearrange this integral

It becomes:

$-R^2 \int_\frac{\pi}{2}^{0} \sqrt{1- cos(\theta)^2}sin(\theta))d\theta$

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AstroGuy

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Thay makes sense

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But there is an identity that says:

sqrt(1-cos^2)=sin(theta)

So we get:

$-R^2\int_\frac{\pi}{2}^{0} sin(\theta)^2d\theta$

[manic canyon]

AstroGuy

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Much simpler

More beautiful

[strange belfry]

Yep

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And what is the identity for sin^2

You might know it

[strange belfry]

1-cos²theta

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$sin(\theta)^2 = \frac{1 -cos(2\theta)}{2}$

[manic canyon]

AstroGuy

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This one

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Oh

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Plug that in your integral

$-R^2 \int_\frac{\pi}{2}^{0} \frac{1 -cos(2\theta)}{2}d\theta$

[manic canyon]

AstroGuy

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There

You can solve that now

I count on you

It’s like a puzzle

[strange belfry]

Alr

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You can get the 1/2 out

Because it’s a constant

[strange belfry]

Yeah

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And the negative sign of R^2

You can switch the bounds

$\frac{R^2}{2} \int_ 0^{\frac{\pi}{2}} (1 -cos(2\theta))d\theta$

[manic canyon]

AstroGuy

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Here

[strange belfry]

Wiat what do you do with the negative sign on -R²?

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$-\int_{a}^{b} dx = \int_{b}^{a} dx$

[strange belfry]

Wait, switching the bounds negates the integral, right?

Yeah

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AstroGuy

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Yeah that’s what I did

You went from this:

$-R^2 \int_\frac{\pi}{2}^{0} \frac{1 -cos(2\theta)}{2}d\theta$

[manic canyon]

AstroGuy

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To this:

$\frac{R^2}{2} \int_ 0^{\frac{\pi}{2}} (1 -cos(2\theta))d\theta$

[manic canyon]

AstroGuy

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Okay, so I have no idea what to do with the $1-\cos(2\theta)$

[manic canyon]

cool looking building

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Now you integrate

[strange belfry]

u substitution?

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No

Just integrate it

You know how to do it..?

[strange belfry]

Nope 😭💔

I'm 15, I only know small fragments of calculus

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It’s okayyy

Integral of cos(2x) is (1/2)sin(2x)

[strange belfry]

Ohh okay

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So integral of (1-cos(2x)) is..

[strange belfry]

1\theta - 1/2\sin(2\theta)

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Yes

So get me that integral value now

$\frac{R^2}{2} [\theta - \frac{1}{2}sin(2\theta)]_{0}^{\frac{\pi}{2}}$

You agree with that ?

Now one last step

You can do it

[manic canyon]

AstroGuy

[strange belfry]

Yeah I was working it out

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Okay okay my bad

And tell me the final value once you’re done

[strange belfry]

$\frac{R^2\pi}{4}$?

[manic canyon]

cool looking building

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Yes!

Exactly!

because we computed the integral of sqrt(R^2-x^2) from x=0 to x=R

,w plot sqrt(R^2 - x^2) from x=0 to x=R

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That’s what we did

That’s a quarter of a circle

So piR^2/4

To get half a circle you would go from x=-R to x=R

,w plot sqrt(R^2 - x^2) from x=-R to x=R

[manic canyon]

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But since we chose x=0 to x=R we get a quarter of a circle area

Now if you go from theta=0 to theta=2pi

You get pi * R^2

A full circle

[strange belfry]

Yeah

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$\frac{R^2}{2} [\theta - \frac{1}{2}sin(2\theta)]_{0}^{2\pi}$

[manic canyon]

AstroGuy

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Just change the bounds of this

And you get what you want

This gives piR^2

Because the bounds are 0 and 2pi

Did you understand everything so far?

[strange belfry]

Yes

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Yayyyyy

Happy happy happy

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It was a rather long explanation

I know..

[strange belfry]

It's alr

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Anything else I could help you with?

[strange belfry]

Nah, ty bro

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Alright I’ll close this channel

Take care

[strange belfry]

Take care

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And have a good day/afternoon/night

[strange belfry]

You too bro

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