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Describe the set of points Calc III equation of sphere

  1. Aug 31, 2005 #1
    Hello everyone, it was the first day of calc III class and this one problem says: Describe the set of points whose coordinates satisfy the given equation and inequalitites. Sketch the graphs.
    x^2 + y^2 + z^2 = 25; x = 3;

    What exactly am I suppose to do? The book doesn't show any examples of this, they just go straight into saying what an equation of a sphere is with center C(h,k.l) and radius r is (x-h)^2 +(y-k)^2 +(z-l)^1 = r^2; In particular, if the center is orgin O, then an equation of the sphere is
    x^2+y^2+z^2 = r^2;
    Now this looks very smilair to the above equation given, so i'm assuming its a sphere, with radius 5. But whats the x for, and how do i describe the set of points whose coordinates satisfy the given equation?> THanks
  2. jcsd
  3. Aug 31, 2005 #2


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    If it wasn't for the x=3, describing the set would just be saying "the sphere of radius 5 centered at the origin." But they want the points that satisfy the first equation (ie, are on the sphere) AND satisfy x=3. The set of points that satisfy x=3 is a plane. What does the intersection look like?
  4. Aug 31, 2005 #3
    The set of points that satisfy those two equations are points that lie on the line of intersection of the two equations. One is a sphere of radius 5, and the other is a YZ plane at x = 3.

    Hint: The plane is slicing the sphere.. what does the slice of a sphere look like?
  5. Aug 31, 2005 #4
    So the sphere actually gets intersected by a plane? I think the rule is, if k is a constant, then x = k represents a plane parallel to the yz-plane. Right? so how does this help me? sorry i'm kinda lost
  6. Aug 31, 2005 #5
    Imagine R^3, imagine a sphere in R^3 centered about the origin, the sphere'sradius is 5 units, so it stretches out 5 units in each direction. Now theres a plane parallel to the YZ plane at x = 3 just standing there, looking like a wall. Obviously since the plane is only 3 units away from the origin, some parts of it are intersecting the sphere, the question is asking you what the line of intersection looks like.

    Its similar to finding the intersection point of two lines in R^2, except in R^3 intersections dont result in just points, but lines.
  7. Aug 31, 2005 #6
    OKay I think this is what i should have done...
    Is that all i'm supppose to do or did i f it up? Thanks
  8. Aug 31, 2005 #7
    Remember, x=3 is a plane, not just a line. Looks good otherwise.
  9. Aug 31, 2005 #8
    Thanks for the help, i know its a plane but i'm not sure how i can draw that without making a big mess. So with that drawing it describes the points satisfying the equation?
  10. Sep 1, 2005 #9
    No you need to explain the set of points is described as. Look at my hint above.
  11. Sep 1, 2005 #10
    "Describing the points that satisfy two equations" is usually the same thing as setting two equations equal to each other, and a picture will not describe the points with the precision that an equation would. You have

    (1) [tex]x^2+y^2+z^2=25[/tex]
    (2) [tex]x=3[/tex]

    Asking which points are common to both equations in this case is the same as substituting equation (2) into equation (1).

    (3) [tex](3)^2+y^2+z^2=25[/tex]
    (3) [tex]9+y^2+z^2=25[/tex]
    (3) [tex]y^2+z^2=16[/tex]

    This is an equation parallel to the yz plane in the plane x=3, as whozum has been explaining up to this point. I think it's immensely important that you understand what this new equation means! People have repeated that the intersection between a plane and a sphere create....some shape that you're familiar with, but you have not mentioned it, which means you don't yet have your head around it.

    Take an infinitely thin slice of a hollow ball...what shape is it? If you were given the initial equation that described the points on that hollow ball, how do you precisely describe the points on the edge of that shape?

    Changed "in the yz plane" in the second paragraph. That was a mistake pointed out by HallsofIvy.
    Last edited: Sep 1, 2005
  12. Sep 1, 2005 #11


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    Severian596 said what I was about to: you are looking for the points that satisfy both
    [tex]x^2+ y^2+ z^2= 25[/tex]
    (which is obviously the sphere centered at (0,0,0) with radius 5)
    and x= 3 (which is the plane (3,y,z) parallel to the yz-plane and 3 units above it).
    Geometrically, you should be able to see the intersection of those two sets, the set of points that are in both is a circle. Analytically, putting x= 3 into the equation of the circle gives
    [tex]y^2+ z^2= 16[/tex]
    which is a circle of radius 4.
    My point now is be careful!. This is still a figure in R3. It is specifically the circle of radius 4 with center at (3, 0, 0) and in the plane x= 3, parallel to the yz-plane.
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