Surface integral x dS over a conical surface

This problem asks to evaluate a surface integral of the first kind over the part of a conical surface cut off by a cylinder.

Problem

Evaluate the surface integral of the first kind:

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

Solution

Step 1

The cylinder x² + y² = 2x defines the boundary of the region in the Oxy plane. Complete the square:

What does completing the square give?

The equation (x−1)² + y² = 1 describes a circle of radius 1 centred at (1; 0). This is important for determining the integration region in the Oxy plane: it is a disk shifted from the origin.

x^{2} - 2 x + y^{2} = 0 \Rightarrow {(x - 1)}^{2} + y^{2} = 1

Step 2

The surface is given by the graph z = √(x²+y²). The surface-area element is computed by the formula:

Why is dS = √2 dx dy?

For a surface given by z = f(x, y), the surface-area element is dS = √(1 + (∂z/∂x)² + (∂z/∂y)²) dx dy. Here ∂z/∂x = x/√(x²+y²) and ∂z/∂y = y/√(x²+y²). The sum of squares of these derivatives equals 1, so the radicand reduces to 2.

d S = \sqrt{1 + {(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2}} d x d y

\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2} + y^{2}}}, \frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^{2} + y^{2}}}

d S = \sqrt{2} d x d y

Step 3

Change to polar coordinates: x = r cos φ, y = r sin φ, dx dy = r dr dφ. The cylinder equation becomes:

r^{2} = 2 r cos φ \Rightarrow r = 2 cos φ, φ \in [- \frac{π}{2}; \frac{π}{2}]

Why are the limits in φ from −π/2 to π/2?

From r = 2 cos φ it follows that r ≥ 0, so cos φ ≥ 0. This condition holds for φ ∈ [−π/2, π/2]. In these limits cos φ is non-negative and the region D is fully covered.

Step 4

Substitute x = r cos φ and dS = √2 · r dr dφ into the integral:

\iint_{S} x d S = \sqrt{2} \int_{- π / 2}^{π / 2} cos φ d φ \int_{0}^{2 cos φ} r^{2} d r

Step 5

Evaluate the inner integral over r:

\int_{0}^{2 cos φ} r^{2} d r = \frac{{(2 cos φ)}^{3}}{3} = \frac{8 {cos}^{3} φ}{3}

Step 6

After substitution we obtain the integral over φ:

\iint_{S} x d S = \frac{8 \sqrt{2}}{3} \int_{- π / 2}^{π / 2} {cos}^{4} φ d φ

How do I evaluate the integral of cos⁴φ?

The integral ∫_{−π/2}^{π/2} cos⁴φ dφ can be evaluated via the power-reduction formula cos²φ = (1 + cos 2φ)/2. Applying it twice and integrating gives the standard result 3π/8.

Step 7

Substitute the value of the cos⁴φ integral and obtain the final answer:

\int_{- π / 2}^{π / 2} {cos}^{4} φ d φ = \frac{3 π}{8}

\iint_{S} x d S = \frac{8 \sqrt{2}}{3} \cdot \frac{3 π}{8} = π \sqrt{2}

Answer

π√2.

Frequently asked questions

How do I evaluate a surface integral of the first kind?

A surface integral of the first kind ∬_S f(x, y, z) dS reduces to a double integral over the projection D of the surface onto a coordinate plane: dS = √(1 + (∂z/∂x)² + (∂z/∂y)²) dx dy. The resulting double integral is then evaluated by standard methods.

How do I compute the surface element dS for a surface z = f(x, y)?

If the surface is given explicitly as z = f(x, y), the surface element is dS = √(1 + (∂z/∂x)² + (∂z/∂y)²) dx dy. In our case for the cone z = √(x²+y²) this formula gives dS = √2 dx dy — a constant factor.

Why does the factor r appear when changing to polar coordinates?

The factor r is the Jacobian of the change from Cartesian to polar coordinates: dx dy = r dr dφ. Geometrically it reflects the stretching of area under the mapping. Without this factor the integral would be computed incorrectly.

What is the difference between surface integrals of the first and second kind?

A surface integral of the first kind (over surface area) does not depend on the choice of side and contains dS. A surface integral of the second kind (flux) depends on the side and contains projections of dS onto coordinate planes (dy dz, dz dx, dx dy). Our problem is of the first kind.