Find the surface area of revolution about the y-axis of x = √ 49 − y 2 over the interval 0 ≤ y ≤ 3

Question

Find the surface area of revolution about the y-axis of  over the interval 

Answer 1

You're working towards the formula S = ∫2πx ds

where ds = √(1+(dx/dy)² ) dy

Working from right to left:

1. dx/dy = -y/√(49-y²)
2. (dx/dy)² = y²/(49-y²)
3. 1+(dx/dy)² = 49/(49-y²)
4. √(1+(dx/dy)² ) = 7/√(49-y²)
5. S = ∫2πx. 7/√(49-y²) dy
6. Plug in for x: S = ∫2π√(49-y²). 7/√(49-y²) dy
7. S = ∫14π dy
8. S = ∫³₀ 14π dy = 42π