A square is rotated about an axis lying in the plane of the square, which intersects the square only at one of its vertices. For what position of the axis, is the volume of the resulting solid of revolution the largest?
To start off, I assumed the axis passing through one of the four vertices (without loss of generality), and inclined at an angle $\theta$ with one of the sides. Clearly, the axis makes an angle $90-\theta$ with the other side passing through the same vertex. If there exists a maximum, I feel that it should occur at $\theta = 45°$, due to symmetry.
I'd appreciate it if someone could verify the same and help me formulate a more mathematical argument for the same. Please point me in the right direction. Thanks a lot.
I think you want Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area A and the length of the path traced by its centroid R, i.e., 2πR. The bottom line is that the volume is given simply by $V=2\pi RA$.
Since the area is fixed, and the maximum distance of the centroid is at $\theta=45^{\circ}$, then your intuitive solution is correct.