Step 1: Concept Explanation:
This problem involves finding the surface area of a revolution. We are given a curve defined as a function of \(y\), specifically \(X=g(Y)\), and this curve is rotated around the y-axis.
Step 2: Formula and Approach:
The surface area \(S\) obtained by rotating the curve \(x = g(y)\) from \(y=c\) to \(y=d\) about the y-axis is given by:
\[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy \]
Note that \(X = \sqrt{9-Y^2}\) is the right half of the circle \(X^2+Y^2=9\) with a radius of 3. The revolution creates a zone on a sphere.
Step 3: Step-by-Step Solution:
Using \(x\) and \(y\) as standard variables, our curve is \(x = \sqrt{9-y^2}\) for \(-2 \leq y \leq 2\).
First, determine the derivative \(\frac{dx}{dy}\):
\[ x = (9-y^2)^{1/2} \]
\[ \frac{dx}{dy} = \frac{1}{2}(9-y^2)^{-1/2}(-2y) = \frac{-y}{\sqrt{9-y^2}} \]
Next, calculate \(1 + \left(\frac{dx}{dy}\right)^2\):
\[ 1 + \left(\frac{dx}{dy}\right)^2 = 1 + \left(\frac{-y}{\sqrt{9-y^2}}\right)^2 = 1 + \frac{y^2}{9-y^2} \]
\[ = \frac{(9-y^2) + y^2}{9-y^2} = \frac{9}{9-y^2} \]
Substitute into the surface area integral with limits from \(y=-2\) to \(y=2\):
\[ S = 2\pi \int_{-2}^{2} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy \]
\[ S = 2\pi \int_{-2}^{2} \sqrt{9-y^2} \sqrt{\frac{9}{9-y^2}} dy \]
\[ S = 2\pi \int_{-2}^{2} \sqrt{9-y^2} . \frac{3}{\sqrt{9-y^2}} dy \]
The term \(\sqrt{9-y^2}\) simplifies out:
\[ S = 2\pi \int_{-2}^{2} 3 dy = 6\pi \int_{-2}^{2} dy \]
\[ S = 6\pi [y]_{-2}^{2} = 6\pi (2 - (-2)) = 6\pi (4) = 24\pi \]
Step 4: Solution:
The surface area is \(24\pi\) square units.