# Integral with polar coordinates

#### Chipset3600

##### Member
Hellow MHB, i'm trying to understand how can i pass this integral to polar coordinates. My biggest doubt is about the "radius".

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hellow MHB, i'm trying to understand how can i pass this integral View attachment 751 to polar coordinates. My biggest doubt is about the "radius".
Hey Chipset3600!

Let's start with the first octant, where the angle $\theta$ is between 0 and $\frac \pi 4$.
It means we have a triangle with angle $\theta$ and a horizontal side of length 1.
What is the length of the hypotenuse?

That would be the upper bound for the "radius" (in this octant).

#### Chipset3600

##### Member
Hey Chipset3600!

Let's start with the first octant, where the angle $\theta$ is between 0 and $\frac \pi 4$.
It means we have a triangle with angle $\theta$ and a horizontal side of length 1.
What is the length of the hypotenuse?

That would be the upper bound for the "radius" (in this octant).
[TEX]is \sqrt{2} [/TEX]

#### Klaas van Aarsen

##### MHB Seeker
Staff member
[TEX]is \sqrt{2} [/TEX]
It will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.
What if it is less?

#### Chipset3600

##### Member
It will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.
What if it is less?
if the angle be zero the hypotenuse tends to infinity

#### Klaas van Aarsen

##### MHB Seeker
Staff member
if the angle be zero the hypotenuse tends to infinity
If the angle is zero, the hypotenuse is just as long as the horizontal side, which has length 1.

Note that the cosine of the angle is the length of the horizontal side divided by the hypotenuse.
$$\cos \theta = \frac{\text{horizontal side}}{\text{hypotenuse}} = \frac 1 r$$
How does this maximum radius $r$ depend on $\theta$?

#### Chipset3600

##### Member
If the angle is zero, the hypotenuse is just as long as the horizontal side, which has length 1.

Note that the cosine of the angle is the length of the horizontal side divided by the hypotenuse.
$$\cos \theta = \frac{\text{horizontal side}}{\text{hypotenuse}} = \frac 1 r$$
How does this maximum radius $r$ depend on $\theta$?
Sorry i don't understood what you mean :/

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Sorry i don't understood what u mean :/
In polar coordinates, you have a radius $r$ and an angle $\theta$.
The angle $\theta$ is bounded by $0$ and $2\pi$.
The radius $r$ is bounded by $0$ and an upper bound that depends on $\theta$.

The formula I just gave, can be rewritten as $r=\dfrac 1 {\cos\theta}$ if the angle is between $0$ and $\frac \pi 4$ (the first octant).

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Sorry i don't understood what u mean :/
Can you perhaps be a bit more specific about what you do and do not understand?

#### Chipset3600

##### Member
In polar coordinates, you have a radius $r$ and an angle $\theta$.
The angle $\theta$ is bounded by $0$ and $2\pi$.
The radius $r$ is bounded by $0$ and an upper bound that depends on $\theta$.

The formula I just gave, can be rewritten as $r=\dfrac 1 {\cos\theta}$ if the angle is between $0$ and $\frac \pi 4$ (the first octant).

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Can you perhaps be a bit more specific about what you do and do not understand?

So in the 1st octant the maximum radius is 1/cos (theta), why can't i say that will be sqrt(2) ?

#### Petrus

##### Well-known member
Hello,
if I understand correct think like this (I have not yet done polar cordinate but I will give it a try)
if you divide by something smal you will get bigger result so with other words the smaler denominator you got the bigger result you get. but think backwords. if your $$\displaystyle \theta$$ is lower then your r is bigger. As I like Serena said $$\displaystyle \cos\theta=\frac{1}{r}$$ and you wanna max your r that means you want your $$\displaystyle \theta$$ be as low as possible.
I would like that you wait for someone else confirm I am correct cause I have not started yet with polar and this is what I understand.

Regards,

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So in the 1st octant the maximum radius is 1/cos (theta), why can't i say that will be sqrt(2) ?
You are integrating over the unit square, like in the following picture.

In your original integral x varies from 0 to 1 and y also varies from 0 to 1.
In polar coordinates your $\theta$ will vary from $0$ to $\dfrac \pi 2$ as you can see in the picture.
And your r will vary from $0$ to $\dfrac 1 {\cos \theta}$ in the first octant (for a given $\theta$).

You can also see that for a given $\theta$ in the first octant the maximum r is not $\sqrt 2$.