Chipset3600
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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".
Chipset3600 said: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".
is \sqrt{2}I like Serena said: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 said:is \sqrt{2}
I like Serena said:It will only be $\sqrt{2}$ if the angle $\theta=\frac \pi 4$.
What if it is less?
Chipset3600 said:if the angle be zero the hypotenuse tends to infinity
I like Serena said: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 said:Sorry i don't understood what u mean :/
Chipset3600 said:Sorry i don't understood what u mean :/
So in the 1st octant the maximum radius is 1/cos (theta), why can't i say that will be sqrt(2) ?I like Serena said: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?
Chipset3600 said:So in the 1st octant the maximum radius is 1/cos (theta), why can't i say that will be sqrt(2) ?