Problem 8 - 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 \(R\). \(f_{1}(x, y)=2 x^{2}+2 y^{2}+3, f_{2}(x, y)=6-x^{2}-y^{2}\); \(R\) is the disk bounded by \(x^{2}+y^{2}=1\)

Step-by-Step Solution

Answer

The volume is \(\frac{3\pi}{2}\).

Step 1: Identify the Volume Function

The volume function between two surfaces is given by the difference of the two functions, which represents the height between the surfaces. Thus, the volume function is: \ \(V(x, y) = f_2(x, y) - f_1(x, y) = (6 - x^2 - y^2) - (2x^2 + 2y^2 + 3)\). \ Simplifying this, we get: \ \(V(x, y) = 3 - 3x^2 - 3y^2 \).

Step 2: Set Up the Double Integral

To find the volume, we need to integrate the volume function \(V(x, y)\) over the region \(R\). The region \(R\) is a disk described by \(x^2 + y^2 \leq 1\). The double integral is: \ \[ \iint_R (3 - 3x^2 - 3y^2) \, dA \].

Step 3: Convert to Polar Coordinates

Since \(R\) is a circular region, it's convenient to use polar coordinates. Transform \(x\) and \(y\) to polar coordinates: \(x = r\cos\theta\), \(y = r\sin\theta\), and \(dA = r \, dr \, d\theta\). \ The integral becomes: \ \[ \int_0^{2\pi} \int_0^1 (3 - 3r^2) r \, dr \, d\theta \].

Step 4: Evaluate the Inner Integral

Evaluate the integral with respect to \(r\): \ \[ \int_0^1 (3r - 3r^3) \, dr = \left[\frac{3r^2}{2} - \frac{3r^4}{4}\right]_0^1 = \left( \frac{3}{2} - \frac{3}{4}\right) \]. \ Simplifying gives \(\frac{3}{2} - \frac{3}{4} = \frac{3}{4} \).

Step 5: Evaluate the Outer Integral

Now, evaluate the integral with respect to \(\theta\): \ \[ \int_0^{2\pi} \frac{3}{4} \, d\theta = \frac{3}{4} \cdot [\theta]_0^{2\pi} = \frac{3}{4} \cdot 2\pi = \frac{3\pi}{2} \].

Step 6: Final Answer

The volume between the surfaces over the region \(R\) is \(\frac{3\pi}{2}\).

Key Concepts

Volume between Surfaces

Finding the volume between two surfaces involves determining the space enclosed by them over a certain region. For this problem, you have two mathematical surfaces, \(f_{1}(x, y)\) and \(f_{2}(x, y)\), and a defined region \(R\) which is a disk in the \(x, y\)-plane. The height between the surfaces is modeled by the difference in their functions; hence, the volume function is \(V(x, y) = f_2(x, y) - f_1(x, y)\). Simplifying results in the function \(V(x, y) = 3 - 3x^2 - 3y^2\). This form gives us the mathematical representation of the height at any point \((x, y)\) within the region.

• The next step involves integrating this volume function over the specified region \(R\).
• Thus, the problem becomes one of evaluating a double integral.

Doing so will yield the total volume between the surfaces for \(R\). Understanding this concept is key in fields like physics and engineering, where calculating the volume of physical objects or spaces is often required.

Polar Coordinates

Polar coordinates provide an efficient way to manage integrals over circular regions. The region \(R\) is defined by the inequality \(x^2 + y^2 \leq 1\), which clearly forms a circle in Cartesian coordinates. Using polar coordinates, we translate these into polar form:

\(x = r\cos\theta\)

\(y = r\sin\theta\)

\(dA = r \, dr \, d\theta\)

The integral is easier to evaluate because the limits for \(r\) are from 0 to 1, and for \(\theta\), they are from 0 to \(2\pi\). This conversion sets up the problem for a much simpler integration process, reflecting how polar coordinates can simplify calculations involving circles or sectors.

Integration in Calculus

Integration in calculus serves as a powerful tool to calculate accumulated quantities, such as area, volume, and total displacement. In this context, we are using a double integral to calculate the volume of space between two surfaces over a specified region \(R\). The process involves:

• Evaluating the inner integral, which is integrated over the radius \(r\): \[\int_0^1 (3r - 3r^3) \, dr = \left[\frac{3r^2}{2} - \frac{3r^4}{4}\right]_0^1\]
• This yields \(\frac{3}{4}\) as the result of the inner integration step.
• The outer integral, \(\int_0^{2\pi} \frac{3}{4} \, d\theta\), deals with angle \(\theta\).

By evaluating these integrals, we find that the total volume is \(\frac{3\pi}{2}\). Each step of integration helps break down complex problems into manageable parts, showcasing how calculus can be used for precise mathematical modeling.

Mathematical Problem Solving

Solving complex mathematical problems necessitates a strategic approach and a good conceptual understanding of various mathematical tools. For this exercise, the method included:

• Identifying and setting up the appropriate volume function, which simplifies the problem by clearly defining what needs to be integrated.
• Recognizing and efficiently using polar coordinates due to the problem's symmetry, which optimized the integration process.
• Executing the integration process step-by-step to ensure an accurate calculation of the desired volume.

Logical reasoning and selecting the right mathematical techniques play a crucial role in successfully solving calculus problems like this one. It demonstrates the importance of understanding not just how to carry out calculations, but also why particular methods are chosen in different scenarios.