Suppose you revolve the plane region completely about the given line to sweep out a solid of revolution. Describe the solid. Then find its surface area in terms of pi.

The answer is explained below

Step-by-step explanation:

The question is not complete, the complete question is Suppose you revolve the plane region completely about the given line to sweep out a solid of revolution. Describe the solid. Then find its surface area in terms of pi.

i) about the y axis

ii) about the x axis

Also the image is attached below.

i) about the y axis

If it is revolved about the y axis, the solid produced is a cone with a radius of of 4 units (r = 4) and a height of 3 units (h = 3). The surface area of the cone is:

Surface area = πr (r + √(r² + h²)) = π × 4 (4 + √(4² + 3²)) = 4π (4 + √25) = 4π (4 + 5) = 4π × 9 = 36π unit²

ii) about the x axis

From the image attached, If it is revolved about the x axis, the solid produced is a cone with a radius of of 3 units (r = 3) and a height of 4 units (h = 4). The surface area of the cone is:

Surface area = πr (r + √(r² + h²)) = π × 3 (3 + √(4² + 3²)) = 3π (3 + √25) = 3π (3 + 5) = 3π × 8 = 24π unit²

3 is greater then both of those numbers