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.
I was trying:
$$dV = \pi \frac{a^2}{h^2}x^2dx$$ $$V = \int_0^h \pi \frac{a^2}{h^2}x^2dx = \frac{1}{3}\pi a^2*h $$
which is not the correct answer
Let the base circle be $x^2+y^2 =a^2$ and the semicircular rod $x^2+z^2=a^2$. Then, the volume of the tent is 4-times that of the first octant, which can be viewed as stack of right isosceles triangular plates alone the $x$ direction. So, the tent volume is integrated as $$V =4\int_0^a \frac12 y^2(x)dx = 4\int_0^a \frac12 (a^2-x^2)\ dx=\frac43a^3 $$
