1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – PinkyWay
    Jun 12, 2022 at 20:32
  • $\begingroup$ @Invisible - Right, or the so-called disk method $\endgroup$
    – Quanto
    Jun 12, 2022 at 20:33
  • $\begingroup$ 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 $\varphi$... $\endgroup$
    – PinkyWay
    Jun 13, 2022 at 6:11
  • $\begingroup$ 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? $\endgroup$
    – PinkyWay
    Jun 13, 2022 at 6:40
  • $\begingroup$ @Invisible - yes, especially for this volume. Not sure how general it is though $\endgroup$
    – Quanto
    Jun 13, 2022 at 13:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .