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Thread: Find 3D cylinder equations

  1. #1
    Member nautica17's Avatar
    Aug 2009

    Find 3D cylinder equations

    This is a question from my calculus 3 class.

    "Find the equations for the following right circular cylinders. Each cylinder has a radius a and is tangent to two coordinate planes."

    Basically I am given three 3D graphs of cylinders tangent to the 3 different planes. The center points for the different graphs are (a, 0, a), (a, a, 0), and (0, a, a). Basically here is what I have for my cylinder equations:

    (a, 0, a) ... ax^2 + az^2 = a
    (a, a, 0) ... ax^2 + ay^2 = a
    (0, a, a) ... ay^2 + az^2 = a

    I am however a little confused as to whether I square the radius or not? Either way, am I doing it right? I feel something is wrong maybe.

    I've attached a picture of the graphs in my textbook.
    Attached Thumbnails Attached Thumbnails Find 3D cylinder equations-img_0725.jpg  
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  2. #2
    Senior Member
    Dec 2010
    A cylinder parallel to an axis just means that the value does not depend on the coordinate along that axis.
    For example, the first cylinder is parallel to the y axis, so you don't need to consider y in the equation. All you need is to write the equation of the circle in the xz plane (this automatically becomes a cylinder in 3 dimensions - think about it).

    So the equation of the circle for the first one is:
    (x-a)^2 + (z - a)^2 = a^2

    Do you understand now?
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  3. #3
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    Start by finding the equation of each circle in the plane...

    The general equation of a circle (in the $\displaystyle \displaystyle x, y$ plane) is $\displaystyle \displaystyle (x - h)^2 + (y - k)^2 = r^2$, where $\displaystyle \displaystyle (h, k)$ is the centre and $\displaystyle \displaystyle r$ is the radius. Though this translates to any of the three coordinate planes.

    I'll do Question 1 and leave 2 and 3 for you...

    In Q.1, the circle is in the $\displaystyle \displaystyle x-z$ plane, so your equation will involve $\displaystyle \displaystyle x$ and $\displaystyle \displaystyle z$.

    The centre is $\displaystyle \displaystyle (a, a)$ and its radius is $\displaystyle \displaystyle a$.

    So the equation of the circle is $\displaystyle \displaystyle (x - a)^2 + (z - a)^2 = a^2$.
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  4. #4
    Member nautica17's Avatar
    Aug 2009
    Ohh, okay I see what to do. Thanks guys! I was looking at the cylinder so hard that I forgot about the circle part. I guess I was looking at a cylinder example at the origin, hence my weird equations.
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