Problem 5 - APEX Calculus

Question

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=8-x^{2}-y^{2}, f_{2}(x, y)=2 x+y ;\) \(R\) is the square with corners (-1,-1) and (1,1) .

Step-by-Step Solution

Answer

The volume is \(\frac{20}{3}\).

Step 1: Identify the Volume to Calculate

We need to calculate the volume between the surfaces described by the functions across the region (square) defined in the x-y plane. This volume is described by the function \( f(x, y) = f_1(x, y) - f_2(x, y) = (8-x^2-y^2)-(2x+y) \).

Step 2: Simplify the Function for Integration

Simplify the integrand to obtain \( f_1(x, y) - f_2(x, y) = 8 - x^2 - y^2 - 2x - y = -x^2 - 2x - y^2 - y + 8 \).

Step 3: Define the Region of Integration

The region \( R \) is a square defined by \(-1 \leq x \leq 1 \) and \(-1 \leq y \leq 1 \). This tells us the bounds of integration for both x and y variables.

Step 4: Setup the Double Integral

The double integral to be evaluated is \[ \int_{-1}^{1} \int_{-1}^{1} (-x^2 - 2x - y^2 - y + 8) \, dy \, dx \].

Step 5: Evaluate the Inner Integral with Respect to y

Performing the inner integration gives us: \[ \int_{-1}^{1} \left[ -x^2y - 2xy - \frac{y^3}{3} - \frac{y^2}{2} + 8y \right]_{y=-1}^{y=1} \, dx \]. Evaluate this: - At \( y = 1 \), it evaluates to \( -x^2 - 2x - \frac{1}{3} - \frac{1}{2} + 8 \).- At \( y = -1 \), it evaluates to \( x^2 + 2x + \frac{1}{3} + \frac{1}{2} - 8 \).

Step 6: Simplify Result of Inner Integral

Simplifying the result from the evaluation, we get: \( 2(-x^2 - 2x) + \frac{16}{3} \).

Step 7: Evaluate the Outer Integral with Respect to x

Now integrate \( \int_{-1}^{1} (-2x^2 - 4x + \frac{16}{3}) \, dx \), which gives:- For \( -2x^2 \), it evaluates to \( -\frac{2}{3} x^3 \).- For \( -4x \), it evaluates to \( -2x^2 \).- For \( \frac{16}{3} \), it evaluates to \( \frac{16}{3}x \).Evaluate all these from \(-1\) to \(1\).

Step 8: Calculate the Definite Integrals

Evaluate the combined results: For \(-\frac{2}{3} x^3\), from \(-1\) to \(1\), result is \(0\).For \(-2x^2\), from \(-1\) to \(1\), result is \( -4 \).For \( \frac{16}{3}x \), from \(-1\) to \(1\), result is \(\frac{32}{3}\).Add these results to get the final volume \( \frac{20}{3} \).

Key Concepts

Double Integrals

In calculus, double integrals are a powerful tool for calculating various properties of two-dimensional areas. Unlike regular integrals that deal with one-dimensional functions, double integrals work with functions over a two-dimensional region. This allows us to find areas, volumes, and other related measurements for surfaces in three-dimensional space.

To set up a double integral, we generally integrate a function over a specified region. The notation usually looks like this: \[\int_{a}^{b} \int_{c}^{d} f(x,y) \, dy \, dx\]This represents integrating the function \( f(x, y) \) first with respect to \( y \) across the limits \( c \) to \( d \), and then with respect to \( x \) from \( a \) to \( b \). The sequence of integration is essential, and it determines the approach to solving the problem.

Double integrals are not just about getting the math right; they provide a conceptual insight into how a quantity is accumulated over an area.

Volume Between Surfaces

When we talk about finding the volume between two surfaces, we're referring to the space that lies vertically between these two surfaces across a designated area in the xy-plane. This is a common task in many fields such as physics and engineering, where understanding the space between surfaces can provide in-depth insights.

The volume between two surfaces defined by \( f_1(x, y) \) and \( f_2(x, y) \) over a specific region is found by taking the difference \( f(x, y) = f_1(x, y) - f_2(x, y) \).

• First, determine the function that describes the region's top surface.
• Next, identify the function for the bottom surface.
• The volume is captured by the double integral of their difference over the region.

This method ensures that you're finding the vertical space between the two layers, excluding any overlapping or voids that do not contain volume. The accuracy of the result depends significantly on properly defining and evaluating the double integral.

Region of Integration

The region of integration is the area over which the double integral is evaluated. It is crucial to establish precise boundaries as they guide the integral calculations. The region is usually given by either explicit coordinates or implicitly by inequalities.

In our exercise, the region \( R \) is defined as a square with corners at \( (-1,-1) \) and \( (1,1) \). This means:

• The x-values range from \(-1\) to \(1\).
• Similarly, the y-values also range from \(-1\) to \(1\).

Such a clear definition simplifies the problem, as it provides straightforward limits for integration. This ensures we consider every point within the intended area while calculating the double integral. Regions can be more complex, and occasionally require switching the order of integration for easier calculations. Making sure the region of integration aligns with the problem’s requirements is a foundational step in solving double integrals.

Steps of Integration

The process of solving a double integral involves multiple steps, each crucial for obtaining an accurate result. Here’s a breakdown of the general progression:

• **Identify the Volume or Function**: Determine the purpose of the integral, such as finding a volume, and create an expression for the function to integrate.
• **Establish the Region of Integration**: Define the specific area over which the integral is evaluated, with clear boundary limits for both variables.
• **Set Up the Double Integral**: Write the integral notation, ensuring that the order (dx and dy) reflects the desired direction of integration.
• **Evaluate the Inner Integral**: Perform the integration for one variable, substituting back any defined limits.
• **Simplify and Evaluate the Outer Integral**: Complete the integration for the remaining variable, also applying any limits.
• **Combine Results**: After both integrations, combine and simplify results to reach your final answer. Make sure to manage all constants and terms accurately.

This process might seem intricate, but by patiently working through each stage, you'll find it becomes manageable. Thoroughly understanding each step helps prevent mistakes and enhances the accuracy of your calculations. Mistakes often arise from missing a limit or incorrectly evaluating a segment, so double-check your workings.