• <blockquote id="a6oys"></blockquote>
  • <tt id="a6oys"></tt>
  • Paul's Online Notes
    Paul's Online Notes
    Home / Calculus II / 3-Dimensional Space / Cylindrical Coordinates
    Show General Notice Show Mobile Notice Show All Notes Hide All Notes
    General Notice

    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

    Mobile Notice
    You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

    Section 6-12 : Cylindrical Coordinates

    As with two dimensional space the standard \(\left( {x,y,z} \right)\) coordinate system is called the Cartesian coordinate system. In the last two sections of this chapter we’ll be looking at some alternate coordinate systems for three dimensional space.

    We’ll start off with the cylindrical coordinate system. This one is fairly simple as it is nothing more than an extension of polar coordinates into three dimensions. Not only is it an extension of polar coordinates, but we extend it into the third dimension just as we extend Cartesian coordinates into the third dimension. All that we do is add a \(z\) on as the third coordinate. The \(r\) and \(q\) are the same as with polar coordinates.

    Here is a sketch of a point in \({\mathbb{R}^3}\).

    This graph has a standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward.  There is a point labeled $\left( x,y,z \right)=\left( r,\theta ,z \right)$ that appears to be in the 1st octant (i.e. x, y, and z are all positive).   From this point a dashed line dropped straight down in the xy-plane (reaching it at a right angle) and the dashed line is labeled “z”.   From the origin a new dashed line is drawn until it reaches the point where the “z” dashed line hits the xy-plane.  The angle from the positive x-axis and the “r” dashed line is shown as $\theta$.

    The conversions for \(x\) and \(y\) are the same conversions that we used back when we were looking at polar coordinates. So, if we have a point in cylindrical coordinates the Cartesian coordinates can be found by using the following conversions.

    \[\begin{align*}x & = r\cos \theta \\ y & = r\sin \theta \\ z & = z\end{align*}\]

    The third equation is just an acknowledgement that the \(z\)-coordinate of a point in Cartesian and polar coordinates is the same.

    Likewise, if we have a point in Cartesian coordinates the cylindrical coordinates can be found by using the following conversions.

    \[\begin{align*}r & = \sqrt {{x^2} + {y^2}} \hspace{0.5in}{\mbox{OR}}\hspace{0.5in}{r^2} = {x^2} + {y^2}\\ \theta & = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\\ z & = z\end{align*}\]

    Let’s take a quick look at some surfaces in cylindrical coordinates.

    Example 1 Identify the surface for each of the following equations.
    1. \(r = 5\)
    2. \({r^2} + {z^2} = 100\)
    3. \(z = r\)
    Show All Solutions Hide All Solutions
    a \(r = 5\) Show Solution

    In two dimensions we know that this is a circle of radius 5. Since we are now in three dimensions and there is no \(z\) in equation this means it is allowed to vary freely. So, for any given \(z\) we will have a circle of radius 5 centered on the \(z\)-axis.

    In other words, we will have a cylinder of radius 5 centered on the \(z\)-axis.


    b \({r^2} + {z^2} = 100\) Show Solution

    This equation will be easy to identify once we convert back to Cartesian coordinates.

    \[\begin{align*}{r^2} + {z^2} & = 100\\ {x^2} + {y^2} + {z^2} & = 100\end{align*}\]

    So, this is a sphere centered at the origin with radius 10.


    c \(z = r\) Show Solution

    Again, this one won’t be too bad if we convert back to Cartesian. For reasons that will be apparent eventually, we’ll first square both sides, then convert.

    \[\begin{align*}{z^2} & = {r^2}\\ {z^2} & = {x^2} + {y^2}\end{align*}\]

    From the section on quadric surfaces we know that this is the equation of a cone.

    亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>