Volume from rotating a functional relationship around a line?

Question

I know how to rotate functions around lines, but a friend of mine proposed that what if I were to be presented a functional relationship that cannot be simplified, and asked what I would do in this case?

As an example, he proposed that I find the volume of the solid of revolution that arises when I rotate the relationship $x^2 + \sin(xy)+y^2=1$ around the line $x=2$.

I considered parameterizing the relationship and somehow integrating it that way, but this has come to nowhere.

Any thoughts at how I would approach this?

Answer 1

I really doubt this can be solved analytically so I will approach this numerically.

Here is the shape in question

Recall that the volume of a solid of revolution generated by a planar shape $L$ is given by the area of $L$ times the distance $D$ the centroid of $L$ travels, as per Pappus's centroid theorem.

Hence, we just need to find these two components. Finding the area of $L$ is not hard, and we can do as follows. Given that $\mathbf{1}_{\chi}(x, y)$ is the indicator function of the set $\chi$ in $\mathbb{R}^2$, the area of $L$ is $$ A=\iint_{\mathbb{R}^2}\mathbf{1}_{x^2 + \sin(xy)+y^2\le1}(x, y)\text{ d}A$$ Accordingly, in mathematica, we set up a notebook like this

 In[1]:= ineq = x^2 + Sin[x*y] + y^2 <= 1
Out[1]:= x^2 + Sin[x*y] + y^2 <= 1
 In[2]:= NIntegrate[Boole[ineq], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, Method -> "LocalAdaptive", AccuracyGoal -> 8, PrecisionGoal -> 8]

This should output

Out[2]:= 3.5123
 In[3]:= NumberForm[%2, 16]
Out[3]:= 3.512296577452417

Now for the centroid, common sense through inspection should show that it is exactly at the origin due to the symmetries of the shape. However, we can prove this as well. We don't care about the $y$ coordinate, since the revolution of the centroid around $x=2$ only depends on the $x$ coordinate.

The centroid's $x$ coordinate is given by $$x_c = \frac{\displaystyle\iint_{\mathbb{R}^2}x\cdot\mathbf{1}_{x^2 + \sin(xy)+y^2\le1}(x, y)\text{ d}A}{\displaystyle\iint_{\mathbb{R}^2}\mathbf{1}_{x^2 + \sin(xy)+y^2\le1}(x, y)\text{ d}A} = \frac{\displaystyle\iint_{\mathbb{R}^2}x\cdot\mathbf{1}_{x^2 + \sin(xy)+y^2\le1}(x, y)\text{ d}A}{A}$$

Finding the top integral in mathematica returns $0$ as expected

 In[4]:= NIntegrate[x*Boole[ineq], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, Method -> "LocalAdaptive", AccuracyGoal -> 8, PrecisionGoal -> 8]
Out[4]:= 0.

Since we revolve our shape $L$ in a circle, the centroid must also travel in a circle. The distance $L$'s centroid travel is hence $2\pi R$, where $R$ is the distance from the line $x=2$ to $x_c$, which is just $2$.

So, the total volume of your solid of revolution is nothing but (approximately) $$2\pi\cdot2\cdot A = 4\pi\cdot 3.512296577452417 = 44.13682...$$