From George Simmons' Calculus With Analytical Geometry, page 229 question 7:
A tent consists of canvas stretched from a circular base of radius $a$ to a vertical semicircular rod fastened to the base at the ends of a diameter. Find the volume of this tent.
Answer (from pg. 864): $\frac{4}{3}a^3$
At first I thought the question meant that the tent was half a sphere, but the answer and looking at the question further show that's not the case. I am having a difficult time just visualizing this. How can the rod be both vertical and attached to the base at two points?
Any insight would be appreciated.
Edit: I think I understand the shape now. Along the diameter where the rod touch the base the shape the shape of the cross-section perpendicular to the base will be semi-circular. Along the perpendicular diameter of the tent, the cross-section perpendicular to the base will be linear. The cross sections in between are some intermediate shape. That said, I've still not completely convinced myself nor am I sure of the calculation even if this is correct.
Ok, I figured it out.
Considering just one half of the tent, the canvas can be sectioned into various triangular wedges of thickness $dx$ as you move from one end of the rod to the other. Both the base and height of each wedge are $\sqrt{a^2 - y^2}$. Thus our triangle has area $\frac{1}{2}(a^2-y^2)$ and thickness $dx$. Taking the integral from $0$ to $a$ and multiplying by $4$ as this integral only includes one quadrant of the circle leads us to our answer.