Problem 25

Question

Show that the surface area of a nonvertical plane over a region $S$ in the $x y$ -plane is $A(S) \sec \gamma$ where $\gamma$ is the acute angle between a normal vector to the plane and the positive $z$ -axis.

Step-by-Step Solution

Verified

Answer

The surface area is scaled by $\sec \gamma$, where $\gamma$ is the angle between the plane's normal vector and the positive $z$-axis.

1Step 1: Understand Plane Equation Setup

A plane in space can be expressed in the form: $ax + by + cz = d$. Here, $(a, b, c)$ is the normal vector to the plane. This normal vector is crucial in finding the angle $\gamma$ with the positive $z$-axis.

2Step 2: Identify Normal Vector and Angle $\gamma$

The normal vector of the plane is $(a, b, c)$. We are interested in finding the angle $\gamma$ between this vector and the $z$-axis, represented as $(0, 0, 1)$. The cosine of the angle between two vectors $(a, b, c)$ and $(0, 0, 1)$ is given by $\cos \gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.

3Step 3: Evaluate $\sec \gamma$

From the cosine relation, $\sec \gamma = \frac{1}{\cos \gamma}$ results in $\sec \gamma = \frac{\sqrt{a^2 + b^2 + c^2}}{c}$. This gives us the factor by which the region $S$ is scaled on the plane.

4Step 4: Calculate Projected Surface Area

The formula for the surface area of the plane over region $S$ is $A(S) = \int\int_S \frac{1}{c} \sqrt{a^2 + b^2 + c^2} \ dA$, where $dA$ is the differential area element in the $xy$-plane. This simplifies to $A(S) \cdot \sec \gamma$.

Key Concepts

Normal VectorAngle With Z-AxisPlane EquationProjected Surface Area

Normal Vector

In the world of mathematics and physics, the concept of a normal vector is essential, especially when dealing with planes and surfaces in three-dimensional space. A normal vector of a plane is a vector that is perpendicular to every point on the plane. This can be visualized as an arrow sticking straight out from the surface of the plane.

In the equation of a plane: $ax + by + cz = d$, the normal vector is represented by the coordinates $(a, b, c)$.
This vector highlights the orientation of the plane in space.
The normal vector plays a crucial role in understanding the geometric properties of the plane, especially when calculating angles with other vectors or axes.

Understanding the normal vector is the first step in exploring how a plane sits in three-dimensional space and interacts with other directions, such as the positive $z$-axis.

Angle With Z-Axis

The angle between the normal vector of a plane and the z-axis can provide insight into how the plane is tilted in 3D space. Understanding this angle, denoted as $\gamma$, involves concepts from vector mathematics.
Finding $\gamma$ helps in understanding the orientation of the plane relative to the z-axis. Here's how you calculate this angle:

Given the normal vector $(a, b, c)$, you will compare it with the z-axis vector, represented as $(0, 0, 1)$.
The cosine of the angle $\gamma$ is calculated using the dot product: $\cos \gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$.

This cosine relationship helps you understand how tilted the plane is with respect to the standard vertical direction.

Plane Equation

The equation of a plane is a fundamental representation that describes all the points forming a flat surface in three-dimensional space. This equation is written in the standard form: $ax + by + cz = d$.
Here's what each part represents:

$a, b, c$: These are the coefficients of the variables $x, y, z$, and jointly constitute the normal vector to the plane.
$d$: A constant that adjusts the plane's position relative to the origin.

Incorporating the normal vector, this equation tells us everything we need to know about the plane's alignment in space. Understanding the plane's equation allows us to compute other properties, like the angle with the z-axis.

Projected Surface Area

To find the surface area of a plane over a specific region, we consider the concept of the projected surface area. This involves projecting points from a three-dimensional plane onto a two-dimensional surface, typically the $xy$-plane.
The task at hand is calculating how much area of the region $S$ is being covered by the plane. Here's a simple breakdown:

The area is scaled by a factor determined by the secant of the angle $\gamma$, which is $\sec \gamma = \frac{\sqrt{a^2 + b^2 + c^2}}{c}$.
The formula thus becomes $A(S) \cdot \sec \gamma$, indicating that the original area $A(S)$ is adjusted based on the tilt of the plane.

By understanding projected surface areas, you can tackle problems involving non-vertical planes and their interaction with a given plane.

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