Let ∑ be the cylinder x ^ 2 + y ^ 2 = a ^ 2 at 0

Let ∑ be the cylinder x ^ 2 + y ^ 2 = a ^ 2 at 0

Symmetry by rotation
∫∫x²ds=∫∫y²ds
=(1/2)∫∫(x²+y²)ds
=(1/2)∫∫a²ds
=(a²/2)∫∫1ds
The integrand is 1 and the integral result is the surface area
=(a²/2)*(2πah)
=πa³h