What is the surface area of a closed circular cylinder measuring R=5' H=6' ?

Answer 1

The total surface area of a closed circular cylinder is
2*pi*r*h+2pi*r^2
=2*pi*r(r+h)
=2*3.14*5(5+6)
=2*3.14*5*11
=345.4 sqft

Answer 2

There are two circles (one at each end of the cylinder).
The lateral area is a rectangle that has been wrapped into a cylinder. It's height is 6 ft, and its length is the circumference of the circle.

A = pi * r^2
A = pi * 5^2
A = 25*pi = 78.54 ft^2

C = 2 * pi * r
C = 2 * pi * 5
C = 10 * pi = 31.42 ft

Lateral Area = 31.42 ft * 6 ft = 188.5 ft^2

78.54 ft^2 + 78.54 ft^2 + 118.5 ft^2 = 110 * pi = 345.58 ft^2

I hope this helps!

Answer 3

toatal area = 2 * (area of the circle) + circle circumference * height
= 2 * pi * (d/2)^2 + pi*d*h
= 39.27 + 94.25
=133.52

Answer 4

25x3.14x2+10x6
=157+60
=217

Answer 5

S.A. = 2.π.r² + π.r.h
S.A. = π.r.(2.r + h)
S.A. = 5π.(10 + 6)
S.A. = 80π
S.A. = 251.3 ft²