Question

consider the following drawing of a paper roll. if the length of the roll is 8 inches, what is the total surface area of the paper roll? 2 of 4 questions sa = 2(πr² + πr₁²) + 8πr² in² sa = πr² - πr₁² + 8πr in² sa = 2(πr² - πr₁²) + 16πr + 16πr₁ in² sa = 2(πr² - πr₁²) + 16πr in²

Explanation:

Step1: Calculate the area of the two circular ends

The area of a circle is \(A = \pi r^{2}\). For the two - end rings (the difference between the larger circle and the smaller circle), the area of one ring is \(\pi r^{2}-\pi r_{1}^{2}\), and for two rings, it is \(2(\pi r^{2}-\pi r_{1}^{2})\).

Step2: Calculate the lateral surface area

The lateral surface area of a cylinder is \(A = 2\pi r h\). Here, the "height" \(h\) (length of the roll) is \(8\) inches. The outer lateral surface area is \(2\pi r\times8=16\pi r\), and the inner lateral surface area is \(2\pi r_{1}\times8 = 16\pi r_{1}\). The total lateral surface area is \(16\pi r+16\pi r_{1}\).

Step3: Calculate the total surface area

The total surface area \(SA\) of the paper roll is the sum of the area of the two circular ends and the lateral surface area. So \(SA=2(\pi r^{2}-\pi r_{1}^{2})+16\pi r + 16\pi r_{1}\text{ in}^{2}\)

Answer:

\(SA = 2(\pi r^{2}-\pi r_{1}^{2})+16\pi r+16\pi r_{1}\text{ in}^{2}\) (the third option)