91Ó°ÊÓ

A plane equation is a mathematical expression that defines a flat, two-dimensional surface in three-dimensional space. Generally, a plane equation is expressed in the form:
\[ ax + by + cz = d \]
Here, \( a \), \( b \), and \( c \) are coefficients that determine the orientation of the plane, and \( d \) is a constant that shifts the plane along the normal vector.

In this exercise, the plane equation is given by:
\[ 2x - 3y + z = 6 \]
This equation represents a plane that intersects the x, y, and z-axes at different points, forming a flat surface. The exercise focuses on the portion of this plane lying over a unit square defined by \( 0 \leq x \leq 1\) and \(0 \leq y \leq 1 \). By rearranging the plane equation, we can express \(z\) in terms of \(x\) and \(y\), which is crucial for surface integrals, obtained as:

• \( z = 6 - 2x + 3y \)

A double integral allows us to compute the volume under a surface over a specified rectangular region. It extends the concept of a single integral over a one-dimensional interval to two-dimensional areas, integrating over both \(x\) and \(y\)-axes. To solve the surface integral \( \iint_{S} g \ dS \) over the region \( R \), we convert it into a double integral:

• Choose limits of integration based on the region \( R \): \( 0 \leq x \leq 1\), \(0 \leq y \leq 1 \).
• Evaluate the integrand, which in this case becomes \( 6x \sqrt{14} \).

Once set up, the problem requires integrating first over one variable (often \(y\)), and then the other (\(x\)), resulting in:
\[ \iint_R 6x \sqrt{14} \ dy \ dx \]
This expression represents the mass density function of the lamina stretched over the unit square on the defined plane.

Partial derivatives measure how a function changes as one of its input variables is modified while keeping the others constant. When dealing with functions of multiple variables, partial derivatives are crucial in examining how the function behaves in each direction. In the context of a surface integral, we often calculate partial derivatives to determine how steep the surface is in the \(x\) or \(y\) direction. For the plane equation \( z = 6 - 2x + 3y \), the partial derivatives with respect to \(x\) and \(y\) are:

• \( \frac{\partial z}{\partial x} = -2 \)
• \( \frac{\partial z}{\partial y} = 3 \)

These values provide information about the tilt of the plane along the \(x\) and \(y\) directions, respectively. They also play a critical role in computing the integrand of the surface integral, specifically to determine the factor under the square root \( \sqrt{1 + 4 + 9} \), which simplifies to \( \sqrt{14} \).

The mass of a lamina is the total sum of its density over a defined surface area. In this exercise, the lamina lies on the plane \( 2x - 3y + z = 6 \), bounded by the unit square. The density function is given by \( g(x, y, z) = xz + 2x^2 - 3xy \), which simplifies to \( 6x \) after substitution and simplification.

The mass is computed using a surface integral, which in this scenario, after converting to a double integral and performing the necessary calculus, results in:

• Integration over the unit square using \( 6x \sqrt{14} \) as the integrand, simplifies the problem.
• The final integral evaluated over the interval \( [0, 1] \) for both \(x\) and \(y\), results in the mass of \( 3\sqrt{14} \).

This yields the mass of the lamina as 11.2248, and by doing so, captures the "weight" of the plane section over the specified area. Understanding this concept is vital for applications in physics and engineering, where planar laminae commonly arise.