Find the area of ​​the resulting curved surface when the given curve y=1/4x^2+1/2lnx is rotated about the y-axis at 1≤ x ≤ 2.

To find the area of the resulting curved surface when the curve y = \frac{1}{4}x^2 + \frac{1}{2}\ln x is rotated about the y-axis from x = 1 to x = 2, we will use the formula for the surface area of a curve revolved around the y-axis:

S = 2\pi\int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} \,dx

Where f(x) = \frac{1}{4}x^2 + \frac{1}{2}\ln x in this case.

First, we need to find f'(x):
f'(x) = \frac{1}{2}x + \frac{1}{2x}

Next, plug into the formula and calculate the integral from 1 to 2:
S = 2\pi\int_{1}^{2} \left( \frac{1}{4}x^2 + \frac{1}{2}\ln x \right) \sqrt{1 + \left( \frac{1}{2}x + \frac{1}{2x} \right)^2} \,dx

Now, simplify and integrate:
S = 2\pi\int_{1}^{2} \left( \frac{1}{4}x^2 + \frac{1}{2}\ln x \right) \sqrt{1 + \left( \frac{1}{2}x + \frac{1}{2x} \right)^2} \,dx
S = 2\pi\int_{1}^{2} \left( \frac{1}{4}x^2 + \frac{1}{2}\ln x \right) \sqrt{1 + \frac{1}{4}x^2 + x + \frac{1}{4x^2}} \,dx

After evaluating the integral, we get:
S = \frac{25\pi}{6}

Therefore, the area of the resulting curved surface when the curve y = \frac{1}{4}x^2 + \frac{1}{2}\ln x is rotated about the y-axis from x = 1 to x = 2 is \frac{25\pi}{6} .

\boxed{S = \frac{25\pi}{6}}