calculus help needed

Question

Find the surface area of revolution for the curve y = √3 x+ 2 revolved about the y-axis for 1 ≤ x ≤ 8

Consider a rectangular swimming pool filled with mercury with surface dimension of 100 ft by 50 ft. Using the 100 ft side as the x-axis with one corner being the origin, the depth of the pool is determined by y = − √ x How much work is required to pump all the mercury out of the pool into a tanker truck three feet above the surface of the liquid? (Mercury weighs 847 lb/cu. ft.)

Answer 1

Use I[a,b] for the integral from a to b and E[a, b] for the evaluation from a to b

dy/dx = √3

I[1,8] 2 π x √(1 + (dy/dx)²) dx =

I[1,8] 2 π x √(1 + (√3)^2) dx =

I[1,8] 2 π x √(1 + 3) dx =

I[1,8] 2 π x * 2 dx =

I[1,8] 4 π x dx =

2 π x² E[1,8] =

2 π (64 - 1) =

126π

We lift the mercury from 3 to √x + 3 feet, as we are lifting 3 feet above the pool. The mercury is 847 pounds and the pool is 50 ft wide.

847 * 50 I[0,100] |[3,√x + 3] y dy dx =

847 * 25 I[0,100] y² E[3,√x + 3] dx =

847 * 25 I[0,100] 9 + 6√x + x - 9 dx =

847 * 25 I[0,100] 6√x + x dx =

847 * 25 (4 x^3/2 + x²/2)E[0,100] =

847 * 25(4000 + 5000) =

847 * 25(9000) =

190575000