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Magazine Chapter 16: Problem 4
Find the area of \(z=x^{2}+y^{2}\) that lies below \(z=1\).
Short Answer
Expert verified
The area is \( \pi \).
Step by step solution
01
Understand the Problem
We are tasked with finding the area of the surface defined by \[ z = x^2 + y^2 \]that lies below the plane\[ z = 1. \]This requires determining the region in the xy-plane that satisfies this geometry.
02
Convert to Polar Coordinates
Since the equation \( z = x^2 + y^2 \) suggests a circle in the xy-plane, it's convenient to switch to polar coordinates, where \( x = r \cos \theta \) and \( y = r \sin \theta \). Thus, \[ z = x^2 + y^2 = r^2. \]
03
Determine Limits in the Polar System
Given \( z = 1 \), the equality \( r^2 = 1 \) determines the boundary condition for \( r \). Hence, \( r = \sqrt{1} \) which gives \( r = 1 \), since the radius can't be negative.
04
Set Up the Integral
In polar coordinates, the area element \( dA \) is given by \( dA = r \, dr \, d\theta \). We integrate over the full circle, so \( r \) ranges from 0 to 1 and \( \theta \) ranges from 0 to \( 2\pi \). Thus, the integral setup is:\[ \int_{0}^{2\pi} \int_{0}^{1} r \, dr \, d\theta. \]
05
Compute the Inner Integral
Evaluate the inner integral with respect to \( r \):\[ \int_{0}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}. \]
06
Compute the Outer Integral
Now evaluate the outer integral, having found the result of the inner integral:\[ \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times \left[ \theta \right]_{0}^{2\pi} = \frac{1}{2} \times 2\pi = \pi. \]
07
Verify and conclude
All computations have been verified correctly, and thus the area of the region where the surface \( z = x^2 + y^2 \) is below \( z = 1 \) is \( \pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates are a two-dimensional coordinate system that specifies the position of a point based on its distance from a reference point (the pole) and the angle from a reference direction. It's especially useful when dealing with problems exhibiting rotational symmetry.
- Polar and Cartesian Conversion: Use connections between Cartesian coordinates (x, y) and polar coordinates (r, 胃):
- For any point, the distance from the origin, or radius, is given by \( r = \sqrt{x^2 + y^2} \).
- The angle, 胃, is measured from the positive x-axis. It is found using \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).
- Conversely, given r and 胃, you can find Cartesian coordinates with \( x = r \cos \theta \) and \( y = r \sin \theta \).
- Application: Used frequently in physics and engineering due to their simplicity when describing circular or radial phenomena
- Ideal for integrating over regions bounded by circles.
- Reduces complexity of solving double integrals with circular symmetry.
Surface Area
Surface area in multivariable calculus refers to the area of a 2-dimensional surface that exists in a 3-dimensional space. This concept is crucial when trying to calculate areas of surfaces defined by functions, especially those that are not easily described using rectangular coordinates.
- Understanding Surface Area: Visualize the surface area like a piece of paper wrapping around a 3D object.
- The formula for surface area involves integrating over the region.
- This can be achieved by setting up an integral in the appropriate coordinates, such as polar or Cartesian.
- Relevance to the Problem:
- The surface defined by \( z = x^2 + y^2 \) forms a paraboloid.
- Finding the region below \( z = 1 \) challenges us to compute the area of this cap-like structure.
- Computational Techniques:
- Incorporates changing variables for efficiency, like using polar coordinates for rotational symmetry.
- Each part of the calculation focuses on slicing the surface into infinitely small sections to calculate the total area.
Double Integral
Double integrals are a way to integrate over a two-dimensional area. They help us to compute quantities like areas, volumes, and average values across surfaces that span more than one dimension.
- Basic Concept: A double integral extends the idea of a single variable integral over a two-dimensional domain.
- Notation can be represented as \( \int \int f(x, y) \, dx \, dy \).
- Allows calculation over surfaces or planes in 3D space.
- Can be visualized as summing up infinitely small products over x and y.
- Application:
- Used to find the total amount of a quantity over a given area.
- In polar coordinates, expressed as \( \int \int f(r, \theta) \, r \, dr \, d\theta \).
- Solving the Exercise:
- Dealing with circular bounds is simplified when using double integrals in polar form.
- Understanding and setting proper limits for each variable guarantees accurate results.
