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

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

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.

the answer is

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

What do i need to do?

Thank you

Re: find an equation for surface of revolution

The equation $\displaystyle x^{2}+z^{2}=[r(y)]^{2}$ is that of a pair of cones (or is it a single cone ?) with a common vertex and an axis of symetry along the $\displaystyle y$ axis. The cones open away from each other from the common vertex. The $\displaystyle r$ in the equation is the radius of the cone and the way that it is written $\displaystyle r(y)$ says that it is a function of $\displaystyle y.$ The radius will be zero when $\displaystyle y=1$ and $\displaystyle 1/2$ when $\displaystyle y=0,$ (or $\displaystyle y=2),$ these coming from the generating line $\displaystyle y=2x+1.$

What you are looking for then is a relationship of the form $\displaystyle r=ay+b$ for some values of $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle r(0)=1/2$ and $\displaystyle r(1)=0.$

That turns out to be $\displaystyle r=(1-y)/2$ and when you substitute that into the given equation it simplifies to your given answer.

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, $\displaystyle x^2+ z^2= ((y- 1)/2)^2= (y^2- 2y+ 1)/4$ so that $\displaystyle 4x^2- y^2+ 4z^2- 2y= 1$.

Re: find an equation for surface of revolution

@HallsofIvy thank you that makes perfect sense.