Problem 314 - Calculus Volume 3

Question

Evaluate surface integral \(\iint_{S} y z d S,\) where \(S\) is plane \(x+y+z=1\) that lies in the first octant.

Step-by-Step Solution

Answer

The surface integral over the first octant plane is zero.

Step 1: Find Equation of the Plane in Terms of Two Variables

We are given the plane equation as \( x + y + z = 1 \). Since the plane lies in the first octant, we can express one variable in terms of the other two. For simplicity, solve for \( z \) in terms of \( x \) and \( y \): \( z = 1 - x - y \).

Step 2: Determine the Limits of Integration

Because the plane is in the first octant, \( x, y, \) and \( z \) must be non-negative. Therefore, \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1-x \) for each fixed \( x \), ensuring \( z = 1 - x - y \geq 0 \).

Step 3: Express the Surface Integral as a Double Integral

The surface integral \( \iint_{S} yz \, dS \) needs to be transformed into a double integral over the region \( R \) on the xy-plane. The integral becomes \( \int_{0}^{1} \int_{0}^{1-x} y(1-x-y) \sqrt{1 + 1 + 1} \, dy \, dx \) because the magnitude of the cross product of the partial derivatives of the plane gives \( \sqrt{3} \) as the differential area element.

Step 4: Set Up and Simplify the Integrand

The integrand is \( yz = y(1-x-y) = y - xy - y^2 \). So the double integral becomes \( \sqrt{3} \int_{0}^{1} \int_{0}^{1-x} (y - xy - y^2) \, dy \, dx \).

Step 5: Integrate with Respect to \( y \)

Perform the inner integration: \(\int_{0}^{1-x} (y - xy - y^2) \, dy \). This results in: \[ \left[ \frac{y^2}{2} - \frac{xy^2}{2} - \frac{y^3}{3} \right]_0^{1-x} = \frac{(1-x)^2}{2} - \frac{x(1-x)^2}{2} - \frac{(1-x)^3}{3}. \]

Step 6: Integrate with Respect to \( x \)

Now, integrate the result over \( x \) from 0 to 1: \( \sqrt{3} \int_{0}^{1} \left( \frac{(1-x)^2}{2} - \frac{x(1-x)^2}{2} - \frac{(1-x)^3}{3} \right) \, dx \). Simplifying and integrating this will yield the final value of the integral.

Step 7: Calculate Final Integral

After simplifying the expressions and performing the integration, the calculated value is:- For \( \frac{(1-x)^2}{2} \): \( \left[ \frac{(1-x)^3}{6} \right]_0^1 \). - For \( \frac{x(1-x)^2}{2} \): Use substitution to integrate if necessary.- For \( \frac{(1-x)^3}{3} \): \( \left[ \frac{(1-x)^4}{12} \right]_0^1 \).Solve these definite integrals and combine them for the result.

Key Concepts

Plane Equation

A plane in three-dimensional space can often be described by a linear equation involving three variables: x, y, and z. In this exercise, the plane is given by the equation: \( x + y + z = 1 \). This equation represents a flat, two-dimensional surface extending infinitely in a three-dimensional space. It is useful to rewrite the plane equation in terms of two variables, especially when calculating integrals. By doing this, you simplify the work by expressing one of the variables in terms of the others. For instance, you can solve for \( z \) as: \( z = 1 - x - y \). This expression helps define the surface when working with integrals.

First Octant

The first octant in three-dimensional space refers to the portion where all the variables \( x, y, \) and \( z \) are non-negative (i.e., \( x \geq 0 \), \( y \geq 0 \), \( z \geq 0 \)). This is essentially a corner of the 3D coordinate system where all coordinate values are positive. It is bounded by the three coordinate planes: the XY, YZ, and ZX planes. For the given plane \( x + y + z = 1 \), only the section that lies in this first octant is of interest because we restrict \( x, y, \) and \( z \) to values that are equal to or greater than zero.

Double Integral

Surface integrals, like the one we are solving, can often be transformed into double integrals. This is possible when the surface in question is expressed in terms of two variables. The double integral is generally computed over a region on the plane (in this case, the xy-plane). For the plane \( S: x + y + z = 1 \) in the first octant, our region of integration R turns into the xy-plane projection. Once the surface integral \( \iint_{S} yz \, dS \) is restructured as a double integral, it becomes: \[ \int_{0}^{1} \int_{0}^{1-x} y(1-x-y) \sqrt{3} \, dy \, dx \] This form allows us to compute the required integral using standard techniques meant for double integrals.

Limits of Integration

Limits of integration define the bounds within which we evaluate our integrals. For a double integral representing a region in space, these limits often vary depending on fixed values of the integrating variables. In the context of the given plane in the first octant, \( x \) ranges from 0 to 1 because these are the feasible values within the introduced plane \( x + y + z = 1 \). For each fixed \( x \), \( y \) will vary from 0 to \( 1-x \), ensuring the constraint \( z = 1 - x - y \geq 0 \) is upheld. Establishing and understanding these limits is crucial to successfully evaluate the integral and derive a meaningful solution.