Implicit equation of revolution $(x^2+y^2)^2=x$

Question

I have a curve given as $(x^2+y^2)^2=x$ in the plane $z=0$. We then rotate this curve around the x axis and then must find a parametrization for the surface as well as find the tangent plane in the point $T\left(\frac{1}{2 \sqrt[3]{2}},-\frac{\sqrt{3}}{2 \sqrt[3]{2}}, 0\right)$.

For the parametrization I used polar coordinates and got: $\vec r(\phi,\alpha) = (\sqrt[3]{cos(\phi)}cos(\phi), \sqrt[3]{cos(\phi)}sin(\phi)cos(\alpha), \sqrt[3]{cos(\phi)}sin(\phi)sin(\alpha) ), \ \phi \in [0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},2\pi], \alpha\in [0,2\pi]$.

Then I saw two routes for solving the second part. Firstly one is obviously computing the partial derivatives and solving for their cross product, however this way would be very tedious. Then I thought about finding the implicit form of the function (as I could then find the gradient) and noticed if I square all the components and add them together I get $\sqrt[3]{cos(\phi)}$. So naturally I would assume the implicit form could be: $x^2+y^2+z^2-\sqrt[3]{cos(\phi)}=0$. But what bothers me about this is that $\sqrt[3]{cos(\phi)}$ is not fixed as it is dependent on $\phi$, so I am not sure this is correct.

So I wonder if this is correct and if so why or why not? Any insight is greatly appriciated.

Answer 1

The cross-section at $x$ has radius $\sqrt{\sqrt x-x^2}$, so the equation of the surface of the revolution is $y^2+z^2=\sqrt x-x^2.$

At $T\left(\frac{1}{2 \sqrt[3]{2}},-\frac{\sqrt{3}}{2 \sqrt[3]{2}}, 0\right)$, the gradient vector of $F(x,y,z)=x^2+y^2+z^2-\sqrt x$ is $\langle 0,-\frac{\sqrt3}{\sqrt[3]2},0\rangle$. So the tangent plane of the surface $F=0$ at $T$ is $y=-\frac{\sqrt3}{2\sqrt[3]2}$.

Answer 2

The curve is defined for $x\ge 0$ with

$$y=\pm \sqrt{\sqrt x - x^2}$$

which requires $0\le x \le 1$ with $y(0)=y(1)=0$.

For the parametrization we could use cylindrical coordinates with respect to $x$ axis, that is

• $x=x \in [0,1]$
• $y=r\cos \theta$
• $z=r\sin \theta$

with $r=|y|=\sqrt{\sqrt x - x^2}$ and $\theta\in [0,2\pi)$.

For the tangent plane at $T$ is suffices to find the tangent line at that point in the plane $z=0$ simply using derivatives for the function $y=-\sqrt{\sqrt x - x^2}$.

Answer 3

First, your parametrization is not quite correct, because your stated interval for $\alpha$ causes the surface to be covered twice, in as much a disk that turns around on a diameter will sweep out a full sphere after only a rotation of $\pi$ radians. You can also simplify your specification of $\phi$ to be $\phi \in (-\pi/2, \pi/2]$.

As for the tangent plane, the fact that the point $T$ has $z = 0$ simplifies the calculation to that of the tangent line through $(x,y) = (2^{-4/3}, -3^{1/2} 2^{-4/3})$. Thus implicit differentiation readily comes to mind:

$$\frac{d}{dx}\left[(x^2+y^2)^2 = x\right] \implies 2(x^2+y^2)\left(2x+2y \frac{dy}{dx}\right) = 1$$ from which we obtain $$\frac{dy}{dx} = \frac{1}{4y(x^2+y^2)} - \frac{x}{y}.$$ Then substitution of the point $T$ into the above gives the slope of such a line, and from the point-slope formula for a line, we get the desired tangent line.