Surface integral z dS over the plane z = 1 inside a circle

This problem asks to evaluate a surface integral of the first kind over a flat disk in a horizontal plane.

Problem

Evaluate the surface integral of the first kind:

\iint_{S} z d S, S : z = 1, x^{2} + y^{2} \leq 1

Solution

Step 1

The surface S is a flat disk of radius 1 in the plane z = 1. The projection of S onto the Oxy plane is the unit disk D: x² + y² ≤ 1.

Why is the projection the unit disk?

The plane z = 1 is parallel to the Oxy plane, so its projection onto Oxy preserves its shape without distortion. The constraint x² + y² ≤ 1 defines the unit disk.

Step 2

Compute the surface element dS for the plane z = 1. The partial derivatives are zero:

\frac{\partial z}{\partial x} = 0, \frac{\partial z}{\partial y} = 0

d S = \sqrt{1 + 0^{2} + 0^{2}} d x d y = d x d y

Step 3

On the surface S the coordinate z is constant and equal to 1, so the integrand simplifies: f(x, y, z) = z = 1.

Why can z be replaced by 1?

On the surface S the equality z = 1 holds for all points. Hence when integrating over this surface the function z can be replaced by the constant value 1. This simplifies the integrand.

Step 4

Pass to a double integral over the projection D:

\iint_{S} z d S = \iint_{D} 1 d x d y

Step 5

The resulting double integral is the area of D. Since D is the unit disk, its area equals π:

\iint_{D} 1 d x d y = π \cdot 1^{2} = π

Answer

π.

Geometric meaning

A surface integral of the first kind of the function f(x, y, z) = 1 over a surface S equals the area of that surface. In our problem the surface is the unit disk, whose area is π. The coincidence of the result with the area of the disk confirms the correctness of the solution.

Frequently asked questions

How do I evaluate a surface integral over a plane?

For a flat surface z = c (where c is a constant), the surface element dS = dx dy, since ∂z/∂x = ∂z/∂y = 0. The integral then reduces to a double integral over the projection D of the surface onto the Oxy plane.

Why does the answer coincide with the area of the disk?

The integrand z = 1 on the whole surface S, and dS = dx dy. Hence the integral becomes the area of the projection D, which coincides with S. The unit disk has area π, so the answer is π.

How do I compute the area of a disk via a double integral?

The area of a region D is computed as ∬_D dx dy. For the unit disk, change to polar coordinates: ∬_D dx dy = ∫₀^{2π} dφ ∫₀^1 r dr = 2π · 1/2 = π. Alternatively, use the known disk-area formula S = πR².

What would change if the plane were z = 3?

If the plane were z = 3, the integrand would become the constant 3, and the result would be 3π. In general, ∬_S z dS for the plane z = c over a region D equals c · area(D).