Calculate Surface Integral of Plane First Octant Chegg

Introduction

The surface integral of a plane in the first octant involves calculating the area of a portion of a plane that lies within the first octant (where x, y, and z are all positive). This is a common problem in physics and engineering when working with surfaces and their properties.

To compute this integral, you'll need to understand the equation of the plane and the limits of integration. The first octant is defined by the region where x ≥ 0, y ≥ 0, and z ≥ 0.

Surface Integral Formula

The general formula for the surface integral of a scalar function f(x,y,z) over a surface S is:

For a plane given by z = ax + by + c, the partial derivatives are:

Substituting these into the formula gives:

For a plane itself (where f(x,y,z) = 1), this simplifies to the area of the plane over the region D.

Calculation Steps

1. Determine the equation of the plane in the form z = ax + by + c.
2. Identify the region D in the xy-plane where the plane is being integrated.
3. Compute the partial derivatives ∂z/∂x and ∂z/∂y.
4. Calculate the integrand √(1 + a² + b²).
5. Set up the double integral over the region D.
6. Evaluate the integral using appropriate techniques (substitution, symmetry, etc.).

Worked Example

Let's calculate the surface integral of the plane z = 2x + 3y over the region D defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1.

1. The plane equation is z = 2x + 3y, so a = 2 and b = 3.
2. The region D is the unit square in the xy-plane.
3. The integrand is √(1 + 2² + 3²) = √(1 + 4 + 9) = √14.
4. The integral becomes ∫∫_D √14 dx dy.
5. Since √14 is constant, the integral evaluates to √14 × area of D = √14 × (1 × 1) = √14.

The surface area of the plane over the unit square is √14.