Consider the graph of an equation graphed in the coordinate plane. [br][br]In calculus, we often end up studying the solid of revolution formed by rotating the graph of such an equation about the Y-AXIS. [br][br]In GeoGebra's 3D Graphing Calculator, this is actually quite easy to do. The silent screencast below illustrates how easy this actually is. [br][br]Here, we illustrate how to EASILY rotate two functions ([math]y=2\sqrt{x}[/math] and [math]y=-2\sqrt{x}[/math] about the Y-AXIS.

However, GeoGebra's Augmented Reality app currently only allows users to plot to surfaces of the form [math]z=[/math]. That is, [i]z[/i] need to be written as a function of [i]x[/i] and [i]y[/i]. [br][br]Yet in order for us to be able to rotate graphs of equations about the Y-AXIS in GeoGebra Augmented Reality, we are restricted, at the moment, to use equations that can be rewritten so that [i]x[/i] is written explicitly as one (or more) function(s) of [i]y[/i]. [br][br]To see this in action, move the [color=#bf9000][b]LARGE YELLOW POINT[/b][/color], in the applet below, up the [color=#38761d]y-axis[/color]. [br]Note how these [b]cross sections [/b]parallel to the xz-plane are always [b]circles with RADIUS = x[/b].

To carry this out in GeoGebra Augmented Reality, we first need to write [i]x[/i] explicitly in terms of [i]y[/i] [br]That is, we need to write [i]x[/i] as a function of [i]y. [/i]Thus, [math]x=f\left(y\right)[/math]. [br][br]Give this, the equation of ANY circular cross section above is [math]x^2+z^2=\left(f\left(y\right)\right)^2[/math][br][br]Upon solving the equation above for [i]z[/i], we obtain [math]z=\sqrt{\left(f\left(y\right)\right)^2-x^2}[/math] and [math]z=-\sqrt{\left(f\left(y\right)\right)^2-x^2}[/math] . [br][br]Thus, given this, any surface of revolution formed by rotating the graph of [math]x=f\left(y\right)[/math] about the Y-AXIS can be considered to be these 2 SURFACES PUT TOGETHER: [br][br][b][color=#1e84cc]z = a surface with POSITIVE OUPUTS (top half)[/color][/b][br][color=#ff00ff][b]z = a surface with NEGATIVE OUTPUTS (bottom half). [br][/b][/color][br][br]Since our original equations (above) were [math]y=\pm2\sqrt{x}[/math], this is equivalent to [math]x=\left(\frac{y}{2}\right)^2[/math]. [br]Thus, [math]x=f\left(y\right)=\left(\frac{y}{2}\right)^2[/math], and we obtain[br] [math]z=\sqrt{\left(\frac{y}{2}\right)^4-x^2}[/math][b][color=#1e84cc]= blue surface shown above. [br][/color][/b][math]z=-\sqrt{\left(\frac{y}{2}\right)^4-x^2}[/math][color=#ff00ff] [b]= pink surface shown above. [/b][/color]