# find an equation for surface of revolution

• Jun 5th 2012, 08:36 PM
icelated
find an equation for surface of revolution
I have a problem where i need to
find the equation for surface of revolution if the generating curve y = 2x + 1, is revolved about the y axis

The book gives this formula for revolving around the y axis

$x^2 + z^2 = [ r(y)]^2$

So, wouldnt i just plug y in and square it?
most of the problems have been done this way

Problem is i dont get the right answer.

$4x^2 -y^2 + 4z^2 + 2y - 1 =0$

What do i need to do?
Thank you
• Jun 6th 2012, 07:35 AM
BobP
Re: find an equation for surface of revolution
The equation $x^{2}+z^{2}=[r(y)]^{2}$ is that of a pair of cones (or is it a single cone ?) with a common vertex and an axis of symetry along the $y$ axis. The cones open away from each other from the common vertex. The $r$ in the equation is the radius of the cone and the way that it is written $r(y)$ says that it is a function of $y.$ The radius will be zero when $y=1$ and $1/2$ when $y=0,$ (or $y=2),$ these coming from the generating line $y=2x+1.$
What you are looking for then is a relationship of the form $r=ay+b$ for some values of $a$ and $b$ such that $r(0)=1/2$ and $r(1)=0.$
That turns out to be $r=(1-y)/2$ and when you substitute that into the given equation it simplifies to your given answer.
• Jun 6th 2012, 10:23 AM
HallsofIvy
Re: find an equation for surface of revolution
BobP, mathematically, its a single cone. Each part, that in "standard English" is called a cone, is a "nappe" of that cone.

icelated, the "r(y)" is the function x= f(y) in the plane. Since you are given the line y= 2x+ 1, x= r(y)= (y- 1)/2. That is what should be squared, $x^2+ z^2= ((y- 1)/2)^2= (y^2- 2y+ 1)/4$ so that $4x^2- y^2+ 4z^2- 2y= 1$.
• Jun 6th 2012, 11:45 AM
icelated
Re: find an equation for surface of revolution
@HallsofIvy thank you that makes perfect sense.