# What is the surface area of the cylinder whose radius is 6 in and the height is 17 in.

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### 2 Answers

Given the cylinder such that the radius of the base is (r) = 6 in

Also, given that the height of the cylinder is (h) = 17 in.

We need to find the surface area of the cylinder.

We know that the surface area of the cylinder is given by the following equation.

SA = 2*area of the base + area of the side.

==> SA = 2*A1 + A2

Let us calculate the area of the base.

We know that the area of the circle = r^2*pi

==> A1 = 36*pi = 113.1 in^2

Now we need to find the area of the sides.

A2 = circumference of the base * h

= 2*pi*r * h

= 2*6*pi*17 = 204*pi = 640.9 in^2

Now we will substitute into the surface area.

SA = 2*A1 + A2

= 2*113.1 + 640.9

= 226.2 + 640.9 = 867.1 in^2 (approx.)

**Then, the surface area of the cylinder is 867.1 in^2**

The surface total area A of the cylinder is given by:

A = 2pirh+pir^2 = 2pir(r+h), where r = radius, h = height. 2pirh is the area of the curved surface of the cylinder and 2pir^2 = is the area of the top and and bottom circular surfaces of the cylinder.

We substitute radius r = 6 inches and height h = 17 inches in A = 2pir(r+h) = 2pi*6(6+17) = 12pi*23 = 276pi = 867.08 sq inches.

Therefore the total surface are of the cylinder = A = 867.07 sq in.