91Ó°ÊÓ

In calculus, the **Jacobian determinant** is a crucial tool when changing variables in multiple integrals. It provides a way to measure how a function transforms volumes in space.
For our surface area problem, we need to find how the surface described by the plane equation \ z = 4 - 2x - y \ scales the infinitesimal area elements in the xy-plane.
Let's break this down into simpler steps:
1. **Partial Derivatives**: We first calculate the partial derivatives of z with respect to x and y:
\( \frac{ \partial z}{\partial x} = -2 \) and \( \frac{\partial z}{\partial y} = -1 \).
2. **Integrand Calculation**: With these derivatives, we find the integrand for the surface area:
\[ \frac{dS}{dx \, dy} = \sqrt{1 + \left( \frac{\partial z}{ \partial x} \right)^2 + \left( \frac{ \partial z}{ \partial y} \right)^2 } \].
Here, the Jacobian essentially helps us transform our integral from the z plane to the xy-plane.

A **double integral** allows us to compute the volume under a surface or the area of a surface in two-dimensional space. In this exercise, it helps to integrate over a region defined by the boundaries of z on the xy-plane.
Here’s how it works:
1. **Set Up the Limits**: We first identify the limits for the variables x and y. According to the problem, we integrate from x = 0 to 1 and y = 0 to 1.
2. **Expression for Surface Area**: The surface area element from the previous section is \( \sqrt{6} dx \, dy \).
3. **Evaluate the Integral**: The double integral is then:
\[ \int_{0}^{1} \int_{0}^{1} \sqrt{6} \, dx \, dy \].
We can separate this into two single integrals since the limits are constants:
\( \sqrt{6} \left( \int_{0}^{1} dx \right) \left( \int_{0}^{1} dy \right) = \sqrt{6} \cdot 1 \cdot 1 = \sqrt{6} \).
The result, \( \sqrt{6} \), gives us the area of the surface.

A **plane equation** in three dimensions is typically given in the form \ Ax + By + Cz = D \. This equation describes a flat surface extending infinitely in all directions within three-dimensional space.
For our exercise, the specific plane equation provided is \ 2x + y + z = 4 \. We can rearrange it to solve for z and describe the height of the plane above the xy-plane:
\( z = 4 - 2x - y \).
The role of the plane equation here is to provide us with a formula to describe the height (z-coordinate) for every point (x, y) in the xy-plane. This height function is crucial for setting up our integral.
By identifying the range of x and y, the equation of the plane helps us define the boundaries for our surface area calculation. The planes \( x=0 \), \( x=1 \), \( y=0 \), and \( y=1 \) bind the region over which we integrate to find the surface area.