The task: find the volume of solid of revolution formed by rotating the area limited by curve $(x+y)^3=axy$ about $x$-axis, where $a$ is a parameter.

How to solve this task?

I tried to convert coordinates to the polar system, but it didn't help.

  • $\begingroup$ Can you add to the question and tell us what you've tried or what your thoughts are on how to proceed? $\endgroup$
    – DMcMor
    Apr 5, 2017 at 13:18
  • $\begingroup$ I tried to convert coordinates to the polar system, but it didn't help. $\endgroup$
    – Byulent
    Apr 5, 2017 at 13:19

1 Answer 1


The polynomial is of third degree and unusable with the straight technique of integrating $\pi y^2\mathrm dx$. By other side, the curve is unbounded, but it has a nice loop in the first quadrant, similar to the Folium of Descartes. I suppose the volume to find is the generated by that loop spining around the $x$-axis. If so, we get some torus-like figure. (All that follows is easily seen if you use a program to plot curves).

We can write the curve in parametric form. Setting $y=xt$, we get:

$$\begin{cases} x(t)=\dfrac{at}{(1+t)^3}\\ y(t)=\dfrac{at^2}{(1+t)^3} \end{cases}$$

Now $\dfrac{\mathrm dx}{\mathrm dt}=\dfrac{a(1-2x)}{(1+t)^4}$

We can now integrate

$$V=\int_{t_i}^{t_f}\pi y^2(t)\dfrac{\mathrm dx(t)}{\mathrm dt}\mathrm dt$$

$$V=\int_{t_i}^{t_f}\pi\left(\frac{at}{(1+t)^3}\right)^2\frac{a(1-2t)}{(1+t)^4}\mathrm dt$$

Now, we need to set $t_i$ and $t_f$. We see the curve crosses itself in $(0,0)$, so, the piece of curve defining our integral is that between the two values of $t$ for wich $x=0$ and $y=0$. These are $t=0$ and $t=\infty$. The integral is not difficult to solve, but cumbersome, so we better let the work to be done by Wolfram Alpha:

$$V=\int_{0}^{\infty}\pi\left(\frac{at}{(1+t)^3}\right)^2\frac{a(1-2t)}{(1+t)^4}\mathrm dt=\frac{-\pi a^3}{420}$$

But the sign is wrong!

Nevertheless we simply get it in positive. If we analize how the curve is drawn as $t$ goes from $0$ forward, we see that first the loop increases $x$, adding volume to the integral but in some point reverses and the loop decreases $x$. The part in reverse overcomes the other part. We have to compute the integral this way:

$$V=\int_{\infty}^{0}\pi\left(\frac{at}{(1+t)^3}\right)^2\frac{a(1-2t)}{(1+t)^4}\mathrm dt=\frac{\pi a^3}{420}$$


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