ELLIPSOIDS

Equation of ellipsoid

An equation that is of the form 
 (((x^2) )/((a^2))) + (((y^2) )/((b^2))) + (((z^2) )/((c^2)))  = 1
is called an ellipsoid.

The equation reminds you of what other figure we have talked about?
   Why do you think that it is called an ellipsoid?

Graph of an ellipsoid

To see what the graph of an ellipsoid will look like, we will use the following ellipsoid for an example:  
            (((x^2))/4) + (((y^2))/9) + (((z^2))/1) = 1.

The following are just programming lines to get the computer to graph the ellipsoid.  You can ignore the programming lines.

Clear[t, u]ParametricPlot3D[{2 Cos[t] Cos[u], 3 Sin[t] Cos[u], Sin[u]}, {t, 
        0, 2 π}, {u, -(π/2), (π/2)}, 
      ViewPoint -> {2.896, 1.244, 1.231}];

Now that you have seen the graph of the ellipsoid, how would you describe it?

Cross-sections of an ellipsoid

If we use a plane and cut the ellipsoid parallel to one of the coordinate planes and then look at the cross-section, we will see some figures that should look familiar to you.
   By the way, your book refers to the cross-section as the trace.  For example, the cross-section you will see when the xy plane cuts the ellipsoid is called the xy-trace.

XY-Trace

To look at the cross section in the xy plane (known as the xy-trace), we need to set z equal to 0 because all the points in the xy plane have a z-coordinate of 0.  The new equation becomes
                (((x^2))/4) + ((y^2)/9) = 1
   This equation should look familiar to you.  The graph of this equation (still in 3-dimensions) is below.

Clear[t, u]ParametricPlot3D[{2 Cos[t] Cos[u], 3 Sin[t] Cos[u], Sin[u]}, {t, 
        0, 2 π}, {u, -(π/2), 0}, 
      ViewPoint -> {2.896, 1.244, 1.231}];

This graph is a little rough, and, although it gives you an idea of what the graph would look like, it might be more interesting to look at it from a different perspective.  If we look at the graph from above, we will see the following:

Clear[t, u]ParametricPlot3D[{2 Cos[t] Cos[u], 3 Sin[t] Cos[u], Sin[u]}, {t, 
        0, 2 π}, {u, -(π/2), 0}, 
      ViewPoint -> {0.000, 0.00, 3.384}];

From this perspective, it is fairly obvious what the shape of the cross-section is.  However, it will be even more obvious if we graph just the outline of the cross-section.

Clear[first, second, x, y, z]first = 
      Plot[(Sqrt[9 - ((9 (x^2))/4)]), {x, -2, 2}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; second = 
      Plot[-(Sqrt[9 - ((9 (x^2))/4)]), {x, -2, 2}, 
        AxesLabel -> {"x", "y"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; Show[first, second, DisplayFunction -> $DisplayFunction];

Remember the equation of the ellipsoid with z = 0 became x¤/4 + y¤/9 = 1.  Can you see why this graph makes sense for this equation?  Of course, math always makes sense.

YZ-Trace

To look at the cross section in the yz plane (known as the yz-trace) of the ellipsoid above, we need to set x equal to 0.  The new equation becomes
                   (((z^2))/1) + ((y^2)/9) = 1
    By now you can tell what the graph of this equation is.  
The graph of this equation is below.

Clear[t, u]ParametricPlot3D[{2 Cos[t] Cos[u], 3 Sin[t] Cos[u], Sin[u]}, {t, 
        (π/2), ((3 π)/2)}, {u, -(π/2), (π/2)}, 
      ViewPoint -> {2.896, 1.244, 1.231}];

Now once again, if we just look at the outline of the cross-section, we will see the following:

Clear[first, second, x, y, z]first = 
      Plot[(Sqrt[1 - ((y^2)/9)]), {y, -3, 3}, 
        AxesLabel -> {"y", "z"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; second = 
      Plot[-(Sqrt[1 - ((y^2)/9)]), {y, -3, 3}, 
        AxesLabel -> {"y", "z"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; Show[first, second, DisplayFunction -> $DisplayFunction];

Once again, look at this graph remembering that its equation is     (((z^2))/1) + ((y^2)/9) = 1

XZ-Trace

To look at the cross section of the ellipsoid in the xz plane (known as the xz-trace), we need to set y equal to 0.  The new equation becomes
                   (((x^2))/4) + ((z^2)/1) = 1
   By now you should definitely know what the graph of this equation will be.  The actual graph of this equation is below.

Clear[t, u]ParametricPlot3D[{2 Cos[t] Cos[u], 3 Sin[t] Cos[u], Sin[u]}, {t, 
        π, 2 π}, {u, -(π/2), (π/2)}, 
      ViewPoint -> {2.896, 1.244, 1.231}];

Once again, the outline of the cross-section is graphed below.

Clear[first, second, x, y, z]first = 
      Plot[(Sqrt[1 - ((x^2)/4)]), {x, -2, 2}, 
        AxesLabel -> {"x", "z"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; second = 
      Plot[-(Sqrt[1 - ((x^2)/4)]), {x, -2, 2}, 
        AxesLabel -> {"x", "z"}, AspectRatio -> Automatic, 
        DisplayFunction -> Identity]; Show[first, second, DisplayFunction -> $DisplayFunction];

Review of Ellipsoids

1)  Which of the following are equations of ellipsoids?
a)   (((x^2))/4) + (((y^2))/81) + (((z^2))/4) = 1
b)   (((x^2))/9) + (((y^2))/25) - (((z^2))/4) = 1
c)    (((x^2))/36) + (y^2) + (((z^2))/25) = 1
d)    4 (x^2) + 9 (y^2) + (z^2) = 36
e)    (((x^2))/9)  + (((z^2))/16) = 1
Answer:

a)  Yes, it is an ellipsoid.
b)  No, it is not an ellipsoid.  The signs must all be positive.
c)  Yes, it is an ellipsoid.
d)  Yes, it is an ellipsoid.  If you divide each side of the equation by 36, it will in the same form that we discussed.
e)  No, it is not an ellipsoid.

2.   For the ellipsoid with the following equation:
               (((x^2))/25) + (((y^2))/16) + (((z^2))/9) = 1
     find the shape of each of the following:
     a)  xy-trace
     b)  yz-trace
     c)  xz-trace
Answer:

a)  ellipse
b)  ellipse
c)  ellipse