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 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.

Answer:

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