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: Find the surface area formula

For a surface $z = f(x,y)$, the surface area is: $$\text{Area} = \iint_D \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} , dx,dy$$

Step 2: Calculate partial derivatives

$$\frac{\partial z}{\partial x} = -2x, \quad \frac{\partial z}{\partial y} = -2y$$

Step 3: Substitute into surface area formula

$$\text{Area} = \iint_D \sqrt{1 + (-2x)^2 + (-2y)^2} , dx,dy$$

$$= \iint_D \sqrt{1 + 4x^2 + 4y^2} , dx,dy$$

$$= \iint_D \sqrt{1 + 4(x^2 + y^2)} , dx,dy$$

Step 4: Determine region $D$

The surface is above $z = 0$: $$16 - x^2 - y^2 \geq 0$$ $$x^2 + y^2 \leq 16$$

So $D$ is the disk of radius 4 centered at the origin.

Step 5: Convert to polar coordinates

Let $x = r\cos\theta$, $y = r\sin\theta$, where $0 \leq r \leq 4$ and $0 \leq \theta \leq 2\pi$.

The Jacobian is $dx,dy = r,dr,d\theta$.

$$\text{Area} = \int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} \cdot r,dr,d\theta$$

$$= \int_0^{2\pi} \int_0^4 r\sqrt{1 + 4r^2} ,dr,d\theta$$

Verification of options:

(A) $\iint_D \sqrt{1 + 4(x^2 + y^2)} , dx,dy$ - Correct

(B) $\iint_D \sqrt{1 + 2(x^2 + y^2)} , dx,dy$ - Incorrect coefficient

(C) $\int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} ,dr,d\theta$ - Missing the $r$ from Jacobian

(D) $\int_0^{2\pi} \int_0^4 \sqrt{1 + 4r^2} ,r,dr,d\theta$ - Correct

Answer: (A) and (D) are correct