Deriving Formulae Related to Circles using Integration
Key Questions

By using polar coordinates, the area of a circle centered at the origin with radius
#R# can be expressed:
#A=int_0^{2pi}int_0^R rdrd theta=piR^2# Let us evaluate the integral,
#A=int_0^{2pi}int_0^R rdrd theta#
by evaluating the inner integral,
#=int_0^{2pi}[{r^2}/2]_0^R d theta=int_0^{2pi}R^2/2 d theta#
by kicking the constant#R^2/2# out of the integral,
#R^2/2int_0^{2pi} d theta=R^2/2[theta]_0^{2pi}=R^2/2 cdot 2pi=piR^2# 
Since a sphere with radius
#r# can be obtained by rotating the region bounded by the semicircle#y=sqrt{r^2x^2}# and the xaxis about the xaxis, the volume#V# of the solid can be found by Disk Method.#V=pi int_{r}^r(sqrt{r^2x^2})^2dx# by the symmetry about the yaxis,
#=2piint_0^r(r^2x^2)dx# #=2pi[r^2xx^3/3]_0^r# #=2pi(r^3r^3/3)# #=4/3pir^3# 
Answer:
Unlike the area, the perimeter of an ellipse cannot easily be written down in closed form, like the formula for the circumference of a circle.
Explanation:
There is a whole area of advanced maths dealing with elliptic integrals. Not only does this give the perimeter of an ellipse as an infinite series, but leads on to solutions for the period of a simple pendulum with large amplitude, [such as here].(http://www.rowan.edu/colleges/csm/departments/math/facultystaff/osler/125%20The%20Perimeter%20of%20an%20Ellipse%20as%20in%20Math%20Sci.pdf)