For some function $f\left(x\right)$ it is possible to rotate it along the x-axis and find the area using
$$\intop_{a}^{b}\pi\left(f\left(x\right)\right)^{2}dx$$
I'm curious how to do this with multivariate calculus. If I represent this in a parametric form, like
$$g\left(t\right)=\left(t,f\left(t\right),0\right)$$
and then rotate using
$$\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & cos\:\theta & -sin\:\theta\\ 0 & sin\:\theta & cos\:\theta \end{array}\right]$$
The result is
$$\varPsi\left(t,\theta\right)=\left(t,f\left(t\right)\left(cos\:\theta\right),f\left(t\right)\left(sin\:\theta\right)\right)$$
over $a\leq t\leq b$ and $0\leq\theta\leq2\pi$ .
I think to compute the volume, I need to use the double-integral
$$V = \iint\left\langle \begin{array}{cc} \frac{\partial\varPsi}{\partial t} & \frac{\partial\varPsi}{\partial\theta}\end{array}\right\rangle \partial t\partial\theta$$
where I'm using $\left\langle \right\rangle$ for matrix determinant out of simplicity (and leaving out the bounds).
I can find
$$\frac{\partial\varPsi}{\partial t}=\left(1,f'\left(t\right)\left(cos\:\theta\right),f'\left(t\right)\left(sin\:\theta\right)\right)$$
and
$$\frac{\partial\varPsi}{\partial\theta}=\left(0,f\left(t\right)\left(-sin\:\theta\right),f\left(t\right)\left(cos\:\theta\right)\right)$$
But the problem is that they would get put in a $3\times2$ matrix, so it doesn't make sense to take the determinant.
So I think I am doing something wrong.
You still need $r$ because it can be anywhere inside the surface
$\Psi(t, r,θ)=(t,r\cos\theta,r\sin\theta)$ $$\iiint \left\langle \begin{array}{cc} \frac{\partial\varPsi}{\partial t} &\frac{\partial\varPsi}{\partial r}& \frac{\partial\varPsi}{\partial\theta}\end{array}\right\rangle \,dr \,d\theta\, dt$$
And then the limit of $r = f(t).$
$$\int_a^b\int_0^{2pi}\int_0^{f(t)} r \,dr \,d\theta\, dt$$ $$\int_a^b \pi f^2(t) dt$$