The given curve is rotated about the y-axis. Find the area of the resulting surface.
y = 1/4x^2 − 1/2ln(x), 4 ≤ x ≤ 5
y' = x/2 - 1/(2x)
ds = √(1+y'^2) dx = √(1 + (x/2 - 1/(2x))^2) dx = 1/2 (x + 1/x) dx
so the surface area is
A = ∫2πr ds
where r = x
A = ∫[4,5] 2πx * 1/2 (x + 1/x) dx = 64π/3
To find the area of the surface obtained by rotating the given curve about the y-axis, we can use the formula for the surface area of a solid of revolution.
The formula for the surface area of a solid of revolution obtained by rotating the curve y = f(x) about the y-axis over the interval [a, b] is given by:
A = 2π ∫[a,b] f(x) √(1 + (f'(x))^2) dx
In this case, the given curve is y = 1/4x^2 - 1/2ln(x) and the interval of rotation is from x = 4 to x = 5.
To find the area, we first need to calculate the function's derivative, f'(x). Then we can plug the function, its derivative, and the interval of rotation into the surface area formula and evaluate the integral.
Let's start by finding the derivative of the given function:
y = 1/4x^2 - 1/2ln(x)
Differentiating with respect to x, we get:
dy/dx = 1/2x - 1/2x^(-1)
Simplifying, we have:
dy/dx = 1/2x - 1/2x^(-1)
Next, we can substitute the function and its derivative into the surface area formula:
A = 2π ∫[4,5] (1/4x^2 - 1/2ln(x)) √(1 + ((1/2x - 1/2x^(-1))^2)) dx
Now, we need to calculate this integral. The first step is to simplify the expression under the square root:
(1/2x - 1/2x^(-1))^2 = (1/2x - 1/2x^(-1))(1/2x - 1/2x^(-1))
= 1/4x^2 - 1/4 + 1/4
= 1/4x^2
Simplifying further, we have:
A = 2π ∫[4,5] (1/4x^2 - 1/2ln(x)) √(1 + (1/4x^2)) dx
Now we can integrate the expression:
A = 2π ∫[4,5] (1/4x^2 - 1/2ln(x)) √(1 + (1/4x^2)) dx
We can evaluate this integral using numerical integration methods or software.