Problem 33 - Calculus with Concepts in Calculus

Question

Determine whether \(f\) is continuous on the given region \(R\). \(f(x, y)=\left\\{\begin{array}{ll}1 & \text { for } x^{2}+y^{2} \leq 9 \\ 0 & \text { for } x^{2}+y^{2}>9\end{array}\right.\) \(R\) is the disk \(x^{2}+y^{2} \leq 9\)

Step-by-Step Solution

Answer

\(f\) is continuous inside the region \(R\) but not on the boundary.

Step 1: Understand the Function

The function \( f(x, y) \) takes on a value of 1 when \( x^2 + y^2 \leq 9 \) and a value of 0 when \( x^2 + y^2 > 9 \). Here, the region \( R \) is defined by the disk \( x^2 + y^2 \leq 9 \). This means \( f(x, y) = 1 \) for all points on and inside the disk and 0 for points outside the disk.

Step 2: Check Continuity Inside Region

Inside the region \( R \), i.e., for all points where \( x^2 + y^2 < 9 \), the function \( f(x, y) = 1 \) is constant. A function is continuous when it doesn't have any abrupt changes, and being constant satisfies this condition. Thus, \( f \) is continuous for all points inside the disk where \( x^2 + y^2 < 9 \).

Step 3: Analyze Continuity on Boundary

At the boundary where \( x^2 + y^2 = 9 \), the function is still defined to be 1. However, just outside this boundary, for \( x^2 + y^2 > 9 \), the function changes to 0. This abrupt change along the boundary implies a discontinuity at the boundary.

Step 4: Verify Discontinuity using Limits

To confirm discontinuity at the boundary, consider approaching any boundary point (like where \( x^2 + y^2 = 9 \)) from within the disk and from outside. Within the disk, the value is 1, but from outside, it approaches 0. The limit from outside (approaching a boundary point) is 0, while from inside it is 1. Since the limits are not equal, \( f(x,y) \) is discontinuous on the boundary \( x^2 + y^2 = 9 \).

Step 5: Conclusion on Continuity

Given the analysis, while \( f(x,y) \) is continuous for \( x^2 + y^2 < 9 \), it has a discontinuity at all points on the boundary \( x^2 + y^2 = 9 \), due to the abrupt change in function value from 1 inside to 0 outside.

Key Concepts

multivariable calculus

Multivariable calculus is the extension of calculus in one variable to functions of several variables. Unlike single-variable calculus, where we deal solely with functions of one variable, multivariable calculus uses functions that take multiple inputs and produce a single output. In our exercise, we are dealing with a function \(f(x, y)\), which is dependent on two variables: \(x\) and \(y\).
This branch of calculus is crucial because many practical phenomena depend on multiple factors, so understanding how to work with multivariable functions can be powerful.

• It introduces concepts such as partial derivatives and gradients, which help determine how changes in one of the input variables impact the output.
• Multivariable calculus also deals with multiple integrals that help calculate volumes under surfaces.
• Understanding limits and continuity in multiple dimensions is more complex as it involves evaluating conditions in a multi-axis plane.

In our case, determining the continuity of \(f(x, y)\) involves understanding its behavior across different regions defined in the xy-plane.

disk region

A disk region in mathematics is often defined as a set of points that are contained within a circle, including the points on the boundary of the circle. In the exercise, the disk region is defined by \(x^2 + y^2 \leq 9\).
This means we are considering all points \((x, y)\) such that their distance from the origin is less than or equal to 3 units (since \(\sqrt{9} = 3\)).

• The set of points inside the disk is where \(x^2 + y^2 < 9\), within which our function \(f(x, y)\) remains constant at 1.
• Points on the boundary are exactly those where \(x^2 + y^2 = 9\), making the outline of our disk.

The simplicity and symmetry of a disk region make it a common geometry in exercises involving continuity and differentiability, as it's easy to visually and mathematically interpret its boundaries and inner regions.

boundary discontinuity

A boundary discontinuity occurs when a function displays a sudden change in its value at the border of a defined region. In our problem, this phenomenon happens precisely at the points on the boundary of the disk defined by \(x^2 + y^2 = 9\). Here, \(f(x, y)\) experiences an abrupt shift from 1 to 0 as you cross the boundary.

• Within the boundary (inside the disk), the function maintains its value of 1.
• Outside the boundary, the function immediately changes to 0.

This is why it’s termed as a discontinuity: because the path from a point inside the disk across its boundary does not have a continuous, smooth transition in value.To mathematically verify this discontinuity, we analyze limits:

• Approaching the boundary from inside the disk, we find the limit is 1, as the function value within is constant at 1.
• Conversely, approaching the disk from the outside, the limit tends towards 0.

Since these two limits do not match, we conclude a discontinuity exists at the boundary.