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Evaluate the surface integral.

∫∫S y dS
S is the part of the paraboloid y = x² + z² that lies inside the cylinder x²+ z² = 9.

User JohnMcG
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1 Answer

5 votes

The value of the surface integral is 3π/5.

Here's how to evaluate the surface integral:

1. Parameterization:

Since the surface S is defined by two equations, we need a parameterization to express x, y, and z in terms of two parameters.

We can use cylindrical coordinates:

θ: angle between the positive x-axis and the projection of the position vector onto the xy-plane (ranging from 0 to 2π)

r: radial distance from the origin (ranging from 0 to 3, as the cylinder has radius 3)

Therefore, the parameterization is:

x = r cos(θ)

y = r² (valid for the paraboloid)

z = r sin(θ)

2. Surface element:

The surface element in cylindrical coordinates is:

dS = r |∂z/∂θ ∂x/∂r - ∂x/∂θ ∂z/∂r| dr dθ

In our case, the partial derivatives are:

∂z/∂θ = r cos(θ)

∂x/∂r = cos(θ)

∂x/∂θ = -r sin(θ)

∂z/∂r = sin(θ)

Therefore, the surface element becomes:

dS = r |r cos²θ + r sin²θ| dr dθ = r² dr dθ

3. Integral setup:

We want to integrate over the area S defined by the cylinder and paraboloid:

∫∫S y dS = ∫₀²π ∫₀³ r² dr dθ

4. Substitution for y:

Substitute y = r² into the integral:

∫₀²π ∫₀³ r⁴ dr dθ

5. Evaluation:

Integrate with respect to r:

∫₀²π ∫₀³ r⁵/5 dr dθ = ∫₀²π 3⁶/30 dθ = 3²/10 ∫₀²π dθ

Integrate with respect to θ:

3²/10 * 2π = 3π/5

Therefore, the value of the surface integral is 3π/5.