Problem 17 - Thomas Calculus

Question

Find the average height of the paraboloid \(z=x^{2}+y^{2}\) over the square \(0 \leq x \leq 2,0 \leq y \leq 2\)

Step-by-Step Solution

Answer

The average height of the paraboloid over the square is \(\frac{8}{3}\).

Step 1: Set Up the Problem

The goal is to find the average height of the given paraboloid. To do this, we need to integrate the function \(z = x^2 + y^2\) over the region \([0, 2] \times [0, 2]\) and then divide by the area of this region. This can be expressed as \( \text{Average Height} = \frac{1}{A} \int \int_R (x^2 + y^2) \, dA \), where \(R\) is the given square and \(A\) is its area.

Step 2: Calculate the Area of the Region

The area of the square \([0, 2] \times [0, 2]\) can be found by multiplying the length of its sides. Thus, \(A = 2 - 0 \times 2 - 0 = 4\).

Step 3: Set Up the Double Integral

To find the integral of \(x^2 + y^2\) over the region \([0, 2]\) for both \(x\) and \(y\), we set up the double integral as \( \int_0^2 \int_0^2 (x^2 + y^2) \, dx \, dy\).

Step 4: Integrate with Respect to \(x\)

First, integrate \(x^2 + y^2\) with respect to \(x\):\[ \int_0^2 (x^2 + y^2) \, dx = \int_0^2 x^2 \, dx + \int_0^2 y^2 \, dx. \]Calculate each part separately:For \(\int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3}\).For \(\int_0^2 y^2 \, dx = y^2(x) \Big|_0^2 = 2y^2\).Thus, the result is \(\frac{8}{3} + 2y^2\).

Step 5: Integrate with Respect to \(y\)

Next, integrate the resulting expression with respect to \(y\):\[ \int_0^2 \left( \frac{8}{3} + 2y^2 \right) \, dy = \int_0^2 \frac{8}{3} \, dy + \int_0^2 2y^2 \, dy. \]Calculate each part separately:For \(\int_0^2 \frac{8}{3} \, dy = \frac{8}{3}y \Big|_0^2 = \frac{16}{3}\).For \(\int_0^2 2y^2 \, dy = \left[ \frac{2y^3}{3} \right]_0^2 = \frac{16}{3}\).Thus, the total integral result is \(\frac{16}{3} + \frac{16}{3} = \frac{32}{3}\).

Step 6: Calculate the Average Height

Divide the result of the double integral by the area \(A = 4\):\[ \text{Average Height} = \frac{\frac{32}{3}}{4} = \frac{32}{12} = \frac{8}{3}. \]

Key Concepts

Double integral

A double integral is a powerful mathematical tool used to compute the integral of a function over a two-dimensional region. It allows us to calculate quantities like area, volume, and other related concepts by integrating over both the x and y dimensions. In the context of the paraboloid given in the problem, the function \(z = x^2 + y^2\) is integrated over a specified square region \([0, 2] \times [0, 2]\). Initially, the integral is set up by assigning limits of integration for both the x and y variables, creating nested integrals.
To solve the double integral, integration is performed in a sequential manner—first integrating with respect to one variable (e.g., x), and then with respect to the other variable (e.g., y). This process allows us to effectively account for variations in both dimensions, resulting in a cumulative integral value over the entire region. Understanding double integrals is crucial for solving complex geometric and physical problems that involve two-dimensional spaces.

Average height

The concept of average height in mathematical terms involves finding the mean value of a function over a specific region. For the given exercise, we're tasked with finding the average height of the paraboloid. This involves computing the average of all the values of the function \(z = x^2 + y^2\) across the defined square region \([0, 2] \times [0, 2]\).
The formula used to find the average height is:

• \(\text{Average Height} = \frac{1}{A} \int \int_R (x^2 + y^2) \, dA \)

where \(A\) represents the total area of the region \(R\). In this situation, the area \(A\), being a square with side lengths of 2, equals 4. By dividing the total integrated value of the function over this region by the area, we derive the average height. This gives insight into how the function behaves on average across the specified area.

Integration over a region

Integration over a region involves calculating the integral of a given function over a delineated area or domain. In this exercise, the region is a square, which simplifies calculations by having straightforward bounds. The limits for \(x\) and \(y\) are both from 0 to 2, creating an area that is easy to visualize and compute integrals over.
When integrating over regions, especially non-rectilinear ones, it's crucial to set up the limits correctly to ensure you are covering the intended area properly. The integration technique utilized must reflect the shape and boundaries of the region of interest.
In the solution, both the x and y variables are integrated step by step, respecting their respective limits. This ensures that the entire specified region is accounted for in calculating the double integral. As a result, we get accurate measures of quantities like area or, as in this case, average height, as the integration encompasses all parts of the region effectively.