Volume of a solid with a circular base Find the volume of the solid with a circular base of radius 9 and the cross sections perpendicular to the y-axis are squares.
I've never solved a problem like this before. How can I go about setting up the integral for this particular problem?
 A: So because of symmetry, split the solid in half along a diameter. This simplifies things.

the cross sections perpendicular to the y-axis are squares

We can thus slice this solid into vertical squares. Have you ever chopped or seen someone chopped a vegetable like a cucumber like in the video below? It is similar to that.
https://www.youtube.com/watch?v=EXgchTrIgw4
So the side length of each slice would be a chord of the circle parallel to the y-axis. The area of each slice would be the square of that. To find the volume of the half-solid we would simply add up the areas of the slices, which is simply integration. So we can integrate the half-solid like this:
$\int_{0}^{9} (2*\sqrt{81 - x^{2}})^2 dx$
From the Pythagorean theorem and area of a square
And then we multiply the above integral by 2.
Since this is an assignment I assume you are in a calculus class or know how to do integration for this problem. Good luck!
A: There are many possibilities.  Take a look at a solid carved out of a unit sphere as a particular case.
When two planes $\dfrac{z}{x}=2$ intersect a sphere radius $\sqrt 5 $ we have two red great circle intersections and all sections normal to y-axis between these intersection great circles are squares.
The squares stand over $xOy$ plane parallel with an edge on this plane being parallel to $xOz$ plane between ellipse projections as shown in a 3d hand sketch below.
Since square discs are stacked integration should proceed with resp to $y$ axis, with elemental volumes obtained by integration with $dy$.

Parameterization of the spheres radius $\sqrt 5$ centered at origin:
$$P(x,y,z)= (\pm t ,\sqrt5 \sqrt{1-t^2}, 2t ) $$
$$ y= \sqrt5 \sqrt{1-t^2} ; dy =\frac{\sqrt5 t dt}{\sqrt{1-t^2}} $$
$$     dV= 4 t^2. dy =4 t^2\frac{\sqrt5 t dt}{\sqrt{1-t^2}}$$
$$     V = 4 {\sqrt5 }\int _0^1\; t^3 {\sqrt{1-t^2}} dt; \; $$
Hope you can take it from here ( substitution $t^2=u$) scaling radius from $\sqrt5$ to 9 by cubing this ratio for volume multiplication and then doubling it for the $-y$ part of the solid.
