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.
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Sign up to join this communityThe 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.
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:
$$V=\int_{\infty}^{0}\pi\left(\frac{at}{(1+t)^3}\right)^2\frac{a(1-2t)}{(1+t)^4}\mathrm dt=\frac{\pi a^3}{420}$$