On bodies of revolution for $y =(1-x^q)^p$

Question

This question is posted in response to a recent one seeking the volume of $y =(a^{2/3}-x^{2/3})^{3/2}$ rotated about the x-axis. I wondered why people don't seek a more general solution when posed with such problems. Suppose tomorrow's problem was a power different than 2/3, of worse yet, what if there were two different powers? Or if it required the rotation about the y-axis?

So, the question is, what is the volume of revolution of $y =(1-x^q)^p$ about the x,y-axes, for arbitrary (p,q)?

What I tried: The equation in question is in the form a general superconics that I have described on the MSE here. I have expanded that to include bodies of revolution about the x,y-axes using Pappus's $2^{}$ Centroid Theorem. The results are shown in the Answer posted below.

Answer 1

The simplest approach to finding the volume of a body of revolution is through Pappus's $2^{}$ Centroid Theorem. Essentially, the volume of a body of revolution is the area multiplied by the path length of the centroid (call it $R=(R_x,R_y)$, depending on the axis of revolution), Thus, $V=2\pi RA$. Now, we have shown previously that the area under the curve $y =(1-|x|^q)^p$, $x\in[-1,1]$ is given by

$$A=\Psi(p,q)=\frac{2\Gamma(p+1)\Gamma(1+1/q)}{\Gamma(p+1+1/q)}$$

This was readily derived from a formal integration leading in turn to the beta and then gamma functions. For the present problem (positive x only), the area would be half that. Now, we can similarly derive equations for the centroid of the curve, and that turns out to be as follows,

$$R=\frac{1}{2A}\biggr(\Psi(2/q,1,p),\Psi(2p,q)\biggr)$$

So, finally we can determine the volume of revolution of $y =(1-x^q)^p$, $x\in[0,1]$ to be

$$V=\frac{\pi}{2}\biggr(\Psi(2p,q),\Psi(2/q,1,p)\biggr)$$

for revolution about the x and y-axes, respectively.

We have determined the volume for the canonical case. In the referenced link we show how to scale the results to the more general case given by $y =b\biggr(1-(x/a)^q\biggr)^p$.

Answer 2

Assuming that the the curve is $|y| = (1 - |x|^q)^p$ so that it is well-behaved for negative $x$, the expression is even, so by symmetry the volume of revolution around the $x$-axis is $$V = 2 \pi \int_0^1 y^2 \,dx = 2 \pi \int_0^1 (1 - x^q)^{2 p} \,dx .$$ Assuming $q > 0$, the substitution $u = x^q$, $du = q x^{q - 1} \,dx$, transforms the integral to $$V_x = \frac{2 \pi}{q} \int_0^1 (1 - u)^{2 p} u^{-\frac{q - 1}{q}} \,du = \frac{2 \pi}{q}\mathrm{B}\left(2 p + 1, \frac{1}{q}\right) = \boxed{\frac{2 \pi}{q} \frac{\Gamma(2 p + 1) \Gamma\left(\frac{1}{q}\right)}{\Gamma\left(2 p + 1 + \frac{1}{q}\right)}} ,$$ provided that $p > -\frac{1}{2}$ (which is necessary for convergence), and where $\mathrm{B}$ and $\Gamma$ are the beta and gamma functions, respectively.

If we solve for $x$ in terms of $y$, we find that $x = (1 - y^\frac{1}{p})^\frac{1}{q}$, and substituting in the above formula for $V_x$ gives that the volume of the solid rotating the curve $|y| = (1 - |x|^p)^q$ around the $y$-axis is $$V_y = \boxed{2 \pi p \frac{\Gamma\left(1 + \frac{2}{q}\right) \Gamma(p)}{\Gamma\left(1 + \frac{2}{q} + p\right)}}$$

Alternatively, we can use the formula $V_y = 2 \pi \int_a^b x y \,dx$, yielding the formula $$V_y = \frac{4 \pi}{q} \frac{\Gamma(p + 1) \Gamma\left(\frac{2}{q}\right)}{\Gamma\left(p + 1 + \frac{2}{q}\right)} ,$$ which can be seen to coincide with the above formula with two applications of the Gamma function identity $\Gamma(z + 1) = z \Gamma(z)$.