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Integral with polar coordinates

Chipset3600

Member
Feb 14, 2012
79
Hellow MHB, i'm trying to understand how can i pass this integral mhb.png to polar coordinates. My biggest doubt is about the "radius".
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
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
Feb 14, 2012
79
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
Mar 5, 2012
8,887

Chipset3600

Member
Feb 14, 2012
79

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,887
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
Feb 14, 2012
79
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
Mar 5, 2012
8,887
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
Feb 14, 2012
79
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).

- - - Updated - - -



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
Feb 21, 2013
739
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
Mar 5, 2012
8,887
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.

polar_on_unit_square.png

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 $\sqrt 2$.