Question

Let \( S \) be that part of the surface of the paraboloid \( z = 16 - x^2 - y^2 \) which is above the plane \( z = 0 \) and \( D \) be its projection on the xy-plane. Then the area of \( S \) equals

• A. \( \iint_D \sqrt{1 + 4(x^2 + y^2)} \, dx \, dy \)
• B. \( \iint_D \sqrt{1 + 2(x^2 + y^2)} \, dx \, dy \)
• C. \( \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, dr \, d\theta \)
• D. \( \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \, d\theta \)

Hint:

For paraboloids and similar surfaces, always switch to polar coordinates when symmetry is evident.

Answer 1

The Correct Option is A, D

Step 1: Formula for surface area.
For a surface \( z = f(x, y) \), the area is \[ A = \iint_D \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx \, dy. \] Given \( z = 16 - x^2 - y^2 \), \[ \frac{\partial z}{\partial x} = -2x, \frac{\partial z}{\partial y} = -2y. \] So, \[ A = \iint_D \sqrt{1 + 4(x^2 + y^2)} \, dx \, dy. \]

Step 2: Convert to polar coordinates.
\[ x = r \cos \theta, \; y = r \sin \theta, \; dx \, dy = r \, dr \, d\theta. \] Since \( z = 0 \) corresponds to \( r^2 = 16 \), \[ A = \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \, d\theta. \]

Final Answer: \[ \boxed{A = \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \, r \, dr \, d\theta.} \]