The given curve is rotated about the y-axis. Find the area of the resulting surface.

Question

The given curve is rotated about the y-axis. Find the area of the resulting surface.

y=1/4 x2 - 1/2 ln(x), 2lessthan/equal xlessthan/equal 3

please show detailed work, thank you!

Answer 1

For a surface revolved around the y-axis the formula is:

S = ∫2πx*√(1+(f '(x))²) dx

First we must find f '(x)

Deriving the first term you get 2x/4 = x/2

the second term becomes -1/2 * d/dx (ln(x)) = -1/2 * 1/x = -1/(2x)

All together that is x/2 - 1/(2x)

once squared it becomes x²/2 - 1/2 + 1/(4x²) -- Remember to FOIL

Plugging this into the radical you get 1+ x²/2 - 1/2 + 1/(4x²) This is from √(1+(f '(x))²)

Combining the like terms 1-1/2 you get: x²/2 + 1/2 + 1/(4x²)

Notice that only the sign changed from -1/2 to 1/2

Next from the formula above S = ∫2πx*√(1+(f '(x))²) dx you must take the square root of

x²/2 + 1/2 + 1/(4x²) This factors into (x/2 + 1/(2x))².

Notice that the sign changed here as well.

Plug this into the main formula you get. S = ∫2πx*√((x/2 + 1/(2x))² )dx

The radical and square root cancel and you're left with ∫2πx*(x/2 + 1/(2x) dx

Simplify: ∫π(x² + 1) dx. (Notice I multiplied the 2x into the parenthesis).

And integrate between the bounds x= 2 to 3

S = π * [x³/3 + x]³₂ The rest is standard integration.

_{S =} π * ((27/3 + 3) - (8/3 + 2)

S = 22π/3