Question

The area of the surface generated by revolving the curve \(X = \sqrt{9-Y^2}, -2 \leq Y \leq 2\) about the y-axis, is

• A. \(24\pi\) Sq. units
• B. \(12\pi\) Sq. units
• C. \(16\pi\) Sq. units
• D. \(48\pi\) Sq. units

Hint:

Recognizing the curve \(x=\sqrt{r^2-y^2}\) as part of a circle can sometimes offer a shortcut using geometric formulas. In this case, the surface is a zone of a sphere of radius 3, with height \(h=2-(-2)=4\). The formula for the area of a spherical zone is \(2\pi rh\), which gives \(2\pi(3)(4) = 24\pi\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the area of a surface of revolution. The curve is given as a function of \(y\) (\(X=g(Y)\)) and is revolved about the y-axis.

Step 2: Key Formula or Approach:
The formula for the surface area \(S\) generated by revolving a curve \(x = g(y)\) from \(y=c\) to \(y=d\) about the y-axis is: \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy \] Note: The curve \(X = \sqrt{9-Y^2}\) represents the right half of a circle \(X^2+Y^2=9\) with radius 3. The surface generated is a zone of a sphere.

Step 3: Detailed Explanation:
Let's use the standard variables \(x\) and \(y\). The curve is \(x = \sqrt{9-y^2}\) for \(-2 \leq y \leq 2\).
First, we find 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, we calculate the term inside the square root in the formula: \[ 1 + \left(\frac{dx}{dy}\right)^2 = 1 + \left(\frac{-y}{\sqrt{9-y^2}}\right)^2 = 1 + \frac{y^2}{9-y^2} \] \[ = \frac{(9-y^2) + y^2}{9-y^2} = \frac{9}{9-y^2} \] Now, we substitute this into the surface area integral. The limits of integration are 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 \(\sqrt{9-y^2}\) cancels 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: Final Answer:
The area of the generated surface is \(24\pi\) square units.