# Difficulty computing volume enclosed by parametrized surface $x=\frac{t^2}{1+t^3},y=\frac {t\cos\phi}{1+t^3},z=\frac {t\sin\phi}{1+t^3}$

Compute the volume of the solid $$V$$ enclosed by the surface $$x(t,\varphi)=\frac{t^2}{1+t^3},y(t,\varphi)=\frac{t}{1+t^3}\cos\varphi,z(t,\varphi)=\frac{t}{1+t^3}\sin\varphi, t\in[0,+\infty),\varphi\in[0,2\pi]$$. You can use the fact that $$\displaystyle\int_0^{+\infty}\frac{dx}{(1+x^3)^2}=\frac{4\pi}{9\sqrt 3}.$$

My attempt:

I think $$V$$ is a rotational solid obtained by rotating a part of the Descartes foil $$\gamma(t):=\left(\underbrace{\frac{t^2}{1+t^3}}_{\gamma_1(t)},\underbrace{\frac{t}{1+t^3}}_{\gamma_2(t)}\right)$$ in the first quadrant around the $$x$$-axis. Since the surface is compact, I think I could apply the divergence theorem for the vector field $$F(x,y,z)=(x,0,0).$$ Let $$\Phi(t,\varphi)=(x(t,\varphi),y(t,\varphi),z(t,\varphi)).$$ Then \begin{aligned}\\&\color{white}=\partial_t\Phi(t,\varphi)\times\partial_\varphi\Phi(t,\varphi)\\&=\begin{vmatrix}\vec i&\vec j&\vec k\\\gamma_1'(t)&\gamma_2'(t)\cos\varphi&\gamma_2'(t)\sin\varphi\\ 0&-\gamma_2(t)\sin\varphi&\gamma_2(t)\cos\varphi\end{vmatrix}\\&=(\gamma_2(t)\gamma_2'(t),-\gamma_1'(t)\gamma_2(t)\cos\varphi,\gamma_1'(t)\gamma_2(t)\sin\varphi).\end{aligned} Just in case if it turned out that computing $$\displaystyle\int_\Phi dA=\int_0^{2\pi}\int_0^{+\infty}\|\partial_t\Phi(t,\varphi)\times\partial_\varphi\Phi(t,\varphi)\|dtd\varphi$$ were an easier way, I computed $$\|\partial_t\Phi(t,\varphi)\times\partial_\varphi\Phi(t,\varphi)\|=\underbrace{\gamma_2(t)}_{\ge 0}\sqrt{\gamma_1'(t)^2+\gamma_2'(t)}=\gamma_2(t)\|\gamma'(t)\|.$$ Now, \begin{aligned}\int_Vdxdydz&=\int_V\operatorname{div}F(x,y,z)dxdydz\\&=\int_\Phi FdA\\&=\int_0^{2\pi}\int_0^{+\infty} F\circ\Phi(t,\varphi)\cdot\left(\partial_t\Phi(t,\varphi)\times\partial_\varphi(t,\varphi)\right)dtd\varphi\\&=2\pi\int_0^{+\infty}\gamma_1(t)\cdot\gamma_2(t)\gamma_2'(t)dt\end{aligned} I thought of substituting $$u=\gamma_2(t)\gamma_2'(t)dt,$$ but $$\gamma_2(t)=\frac{t}{1+t^2}$$ isn't injective on $$[0,+\infty)$$ since $$\gamma_2(0)=0$$ and $$\lim\limits_{t\to+\infty}\gamma_2(t)=0.$$ Is there a more suitable vector field with divergence $$1$$ that I should consider instead or a way to integrate so that I encounter the integral $$\displaystyle\int_0^{+\infty}\frac{dx}{(1+x^3)^2}$$ as expected?

Note that $$y^2+z^2= \frac{t^2}{(1+t^3)^2}$$ and $$dx= \frac{t(2-t^3)}{(1+t^3)^2}dt$$. Then, the volume is given by \begin{align} V= &\int_{x(0)}^{x(\infty)}\pi (y^2+z^2)\ dx =\int_0^\infty \frac{\pi t^2}{(1+t^3)^2} \frac{t(2-t^3)}{(1+t^3)^2}\ dt\\ =& \ \pi \int_0^\infty \left(-\frac{1}{(1+t^3)^2} + \frac{4}{(1+t^3)^3} -\frac{3}{(1+t^3)^4} \right)dt\\ =& \ \pi \left( -1+4\cdot \frac56-3\cdot \frac89\frac56\right) \int_0^\infty \frac{1}{(1+t^3)^2}dt=\frac{4\pi^2}{81\sqrt3} \end{align} where the reduction formula $$\int_0^\infty \frac{dt}{(1+t^3)^n}=I_n = \frac{3n-4}{3n-3}I_{n-1}$$ is applied.
• Does $V=\int_{x(0)}^{x(\infty)}\pi(y^2+z^2)dx$ come from $V=\pi\int_a^b f(x)^2dx$ for the volume of the solid obtained by rotating the graph of $f(x)$ around the $x$-axis? Jun 12, 2022 at 20:32
• I think, if I consider the vector field $F(x,y,z)=(0,-y,2z),$ so that $F\circ\Phi(t,\varphi)=\left(0,-\gamma_2(t)\cos\varphi,2\gamma_2(t)\sin\varphi\right),$ if $N(t,\varphi)$ denotes the normal to the surface, then $$F\circ\Phi(t,\varphi)\cdot N(t,\varphi)=\gamma_1'(t)\gamma_2(t)^2\cos^2\varphi+2\gamma_2'(t)\gamma_2(t)^2\sin^2\varphi=\gamma_1'(t)\gamma_2(t)^2(\cos^2\varphi+2\sin^2\varphi),$$ and then I get the same integral in $t$ variable, but not in $\varphi$... Jun 13, 2022 at 6:11
• Are we allowed to compute the integral the way you did since $y(t)$ isn't a function of $x$ as the curve is closed? Jun 13, 2022 at 6:40