I know that to compute a solid of revolution of a function $f(x)$ rotated around the $y$-axis, one method we can use is the "shell" method. For example, $f(x)=1/4x^2\in [2,4]$, rotated around the $y$-axis would be $\int_2^4{(2\pi x\cdot \frac{1}{4}x^2)dx}$ by shell.
However, how would you compute the volume of solid of revolution whose function is bounded by two variables $x$, and $y$? Take, for example, the region bounded by $(x-8)^2 + y^2 + 14\sin(x-8)+14\sin(y)\le 30$ pictured below.
It is simple enough to compute the area of the shape by using a characteristic function:
$$f(x,y) = \begin{cases} 1, & (x-8)^2 + y^2 + 14\sin(x-8)+14\sin(y)\le 30 \\ 0, & \text{else} \end{cases}$$
$$\text{single area}=\int_{y=-8}^7\int_{x=0}^{15}f(x,y)\ dx\ dy \approx 95.352\color{gray}{72341966629}$$
From this computed area of the figure, is it possible to compute the volume of the solid of the revolution around the $y$-axis of this figure?
Just a thought:
By setting the inequality as an equality and substituting $y=0$,
$(x-8)^2+0+14\sin(x-8)+14\cdot 0=30$
I found the two intersections of the surface at $y=0$ at around $x\approx 3.78962$ and $x\approx 13.9171$.
I wonder if it's possible to calculate the volume by using the same approach that you used to compute the area. However, the indicator function should be $1$ whenever we have the equality $(x-8)^2+y^2+14\sin(x-8)+14\sin(y)=30$.
For the top part, we could use the disk and washers method from $x\approx 3.78962$ to $x\approx 13.9171$ and the portion of $y$ that is above the $x$-axis. However, the problem is that there is that there is that hole in the upper portion and we would have to subtract the volume rotated by that hole.
Also, since you want to rotate this particular figure around the $x$-axis and not the $y$-axis, we have the problem that the volumes of rotation collide (we can't have both the top and bottom portions rotated complete around the $x$-axis. Of course, a simple solution would be to cut the volume in half and let the top portion be a rotation only around the positive $y$-axis and the bottom portion the negative $y$-axis, which we could do by dividing the volume in half.
We could rotate the bottom portion similarly with the proper $x$ and $y$ values.
Just a thought