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#### Chipset3600

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- Feb 14, 2012

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- Thread starter Chipset3600
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- Feb 14, 2012

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- Mar 5, 2012

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Hey Chipset3600!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".

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).

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- Feb 14, 2012

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[TEX]is \sqrt{2} [/TEX]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).

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- #4

- Mar 5, 2012

- 9,806

It will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.[TEX]is \sqrt{2} [/TEX]

What if it is less?

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- Feb 14, 2012

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if the angle be zero the hypotenuse tends to infinityIt will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.

What if it is less?

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- Mar 5, 2012

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

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 $\theta$?

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- #7

- Feb 14, 2012

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Sorry i don't understood what you mean :/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 $\theta$?

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- Mar 5, 2012

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In polar coordinates, you have a radius $r$ and an angle $\theta$.Sorry i don't understood what u mean :/

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 $r=\dfrac 1 {\cos\theta}$

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

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- #9

- 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 $r=\dfrac 1 {\cos\theta}$ifthe 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) ?

- Feb 21, 2013

- 739

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,

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- #11

- Mar 5, 2012

- 9,806

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

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