Question

The area of the part of the surface of the paraboloid \[ x^2 + y^2 + z = 8 \] lying inside the cylinder \[ x^2 + y^2 = 4 \] is

• A. \( \frac{\pi}{2} (17^{3/2} - 1) \)
• B. \( \pi (17^{3/2} - 1) \)
• C. \( \frac{\pi}{6} (17^{3/2} - 1) \)
• D. \( \frac{\pi}{3} (17^{3/2} - 1) \)

Hint:

For surface area integrals over a cylinder, use polar coordinates to simplify the expression for the area element.

Answer 1

The Correct Option is C

Step 1: Parameterize the surface.
We are given the equation of the paraboloid \( x^2 + y^2 + z = 8 \), so \( z = 8 - x^2 - y^2 \). The region of interest is the area inside the cylinder \( x^2 + y^2 = 4 \).
Step 2: Use polar coordinates.
Convert the integral to polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \). The area element in polar coordinates is \( r \, dr \, d\theta \), and the surface area formula becomes: \[ A = \iint_S \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA. \]
Step 3: Set up the integral.
After substituting the expressions for \( z \), compute the integrals, and you will get the result \( \pi (17^{3/2} - 1) \).

Step 4: Conclusion.
Thus, the correct answer is \( \boxed{(B)} \).