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Thread: finding limit of spherical coordinates

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    finding limit of spherical coordinates

    Calculus III - Spherical Coordinates

    <blockquote class="imgur-embed-pub" lang="en" data-id="a/iZORz"><a href="//imgur.com/iZORz"></a></blockquote><script async src="//s.imgur.com/min/embed.js" charset="utf-8"></script>

    [img]http://imgur.com/a/iZORz.jpg[/img]


    Well , i have problem of finding the limit of ρ in this question .

    From the link above , i know that (x^2) + (y^2) + (z^2) = (
    ρ^2)

    In this question , z = 1 and z = sqrt ( (x^2) -(y^2) )

    so , for
    z = sqrt ( (x^2) -(y^2) ) , i have x^2) + (y^2) - (z^2) = 0
    for z = 1 , i cant do nothing.... How to get the limit of
    ρ in this question ? finding limit of spherical coordinates-dsc_0010.jpg
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    Re: finding limit of spherical coordinates

    bump
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    Re: finding limit of spherical coordinates

    Bump, anyone can help??
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    Re: finding limit of spherical coordinates

    $\displaystyle z=\sqrt{x^2+y^2}$

    $\displaystyle z=\sqrt{x^2-y^2}$ ?
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    Re: finding limit of spherical coordinates

    Quote Originally Posted by Idea View Post
    $\displaystyle z=\sqrt{x^2+y^2}$

    $\displaystyle z=\sqrt{x^2-y^2}$ ?
    Sorry, I mean the first one ... So, how to proceed?
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    Re: finding limit of spherical coordinates

    The integral is $\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}}\int_{\sqrt{x^2+ y^2}}^1 dzdydx$.

    Looking at the "outer integral", x goes from x= -1 to x= 1, and then, looking at the second integral, for each x, y goes from $\displaystyle y= -\sqrt{1- x}$ to $\displaystyle y= \sqrt{1- x^2}$. Those last two are the same as $\displaystyle y^2= 1- x^2$ or $\displaystyle x^2+y^2= 1$, the circle with center at (0, 0) in the xy-plane and radius 1. Finally, for every such (x, y), z goes from $\displaystyle z= \sqrt{x^2+ y^2}$, which is the part of a cone with vertex at (0, 0, 0) opening upwards, to z= 1. Are you required to do this in spherical coordinates? I think it would be easier in "cylindrical coordinates": [tex]\int_0^{2\pi}\int_0^1\int_r^1 r dzdrd\theta[tex] since, in cylindrical coordinates, that cone is just z= r.

    If you really must do it in spherical coordinates, then $\displaystyle x= \rho cos(\theta)sin(\phi)$ and $\displaystyle y= \rho sin(\theta) sin(\phi)$ so $\displaystyle \sqrt{x^2+ y^2}= \sqrt{\rho^2cos^2(\theta)sin^2(\phi)+ \rho^2sin^2(\theta)sin^2(\phi)}= \sqrt{\rho^2\sin^2(\phi)(sin^2(\theta)+ cos^2(\theta))}= \rho\sin(\phi)$.

    $\displaystyle z= \rho cos(\phi)$ so $\displaystyle z= \sqrt{x^2+ y^2}$ becomes $\displaystyle \rho cos(\phi)= \rho sin(\phi)$ and $\displaystyle cos(\phi)= sin(\phi)$ or $\displaystyle tan(\phi)= 1$ so that $\displaystyle \phi= \frac{\pi}{4}$ which, now that I think of it, should have been obvious! It is the upper limit, z= 1, that is more complicated. $\displaystyle z= \rho cos(\phi)= 1$ so $\displaystyle \rho= \frac{1}{cos(\phi)}= sec(\phi)$.

    In spherical coordinates, the integral is $\displaystyle \int_{0}^{2\pi}\int_0^{\pi/4}\int_{0}^{sec(\phi)} \rho^2 sin^2(\phi)d\rho d\phi d\theta$
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    Re: finding limit of spherical coordinates

    $\displaystyle \rho =\sec \phi $

    finding limit of spherical coordinates-sphco.jpg
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    Re: finding limit of spherical coordinates

    Quote Originally Posted by HallsofIvy View Post
    The integral is $\displaystyle \int_{-1}^1 \int_{-\sqrt{1- x^2}}^{\sqrt{1- x^2}}\int_{\sqrt{x^2+ y^2}}^1 dzdydx$.

    Looking at the "outer integral", x goes from x= -1 to x= 1, and then, looking at the second integral, for each x, y goes from $\displaystyle y= -\sqrt{1- x}$ to $\displaystyle y= \sqrt{1- x^2}$. Those last two are the same as $\displaystyle y^2= 1- x^2$ or $\displaystyle x^2+y^2= 1$, the circle with center at (0, 0) in the xy-plane and radius 1. Finally, for every such (x, y), z goes from $\displaystyle z= \sqrt{x^2+ y^2}$, which is the part of a cone with vertex at (0, 0, 0) opening upwards, to z= 1. Are you required to do this in spherical coordinates? I think it would be easier in "cylindrical coordinates": [tex]\int_0^{2\pi}\int_0^1\int_r^1 r dzdrd\theta[tex] since, in cylindrical coordinates, that cone is just z= r.

    If you really must do it in spherical coordinates, then $\displaystyle x= \rho cos(\theta)sin(\phi)$ and $\displaystyle y= \rho sin(\theta) sin(\phi)$ so $\displaystyle \sqrt{x^2+ y^2}= \sqrt{\rho^2cos^2(\theta)sin^2(\phi)+ \rho^2sin^2(\theta)sin^2(\phi)}= \sqrt{\rho^2\sin^2(\phi)(sin^2(\theta)+ cos^2(\theta))}= \rho\sin(\phi)$.

    $\displaystyle z= \rho cos(\phi)$ so $\displaystyle z= \sqrt{x^2+ y^2}$ becomes $\displaystyle \rho cos(\phi)= \rho sin(\phi)$ and $\displaystyle cos(\phi)= sin(\phi)$ or $\displaystyle tan(\phi)= 1$ so that $\displaystyle \phi= \frac{\pi}{4}$ which, now that I think of it, should have been obvious! It is the upper limit, z= 1, that is more complicated. $\displaystyle z= \rho cos(\phi)= 1$ so $\displaystyle \rho= \frac{1}{cos(\phi)}= sec(\phi)$.

    In spherical coordinates, the integral is $\displaystyle \int_{0}^{2\pi}\int_0^{\pi/4}\int_{0}^{sec(\phi)} \rho^2 sin^2(\phi)d\rho d\phi d\theta$
    so , there is no way to prove that ρ start from 0 using the formula ? we could only sketch out the 3d diagram and see it ?
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    Re: finding limit of spherical coordinates

    Quote Originally Posted by xl5899 View Post
    so , there is no way to prove that ρ start from 0 using the formula ? we could only sketch out the 3d diagram and see it ?
    ??? I said nothing about drawing a diagram. The fact that the region defined by the original limits of integration includes the origin proves that $\displaystyle \rho$ "starts from 0".
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