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Chipset3600
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- Feb 14, 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".
[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).
It will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.[TEX]is \sqrt{2} [/TEX]
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?
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
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 $\theta$?
In polar coordinates, you have a radius $r$ and an angle $\theta$.Sorry i don't understood what u mean :/
Can you perhaps be a bit more specific about what you do and do not understand?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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Can you perhaps be a bit more specific about what you do and do not understand?
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) ?