Question

Suppose that \(f(x, y) \geq 0\) on \(R=[a, b] \times[c, d]\). Justify the interpretation of \(\int_{R} f(x, y) d(x, y),\) if it exists, as the volume of the region in \(\mathbb{R}^{3}\) bounded by the surfaces \(z=f(x, y)\) and the planes \(z=0, x=a, x=b, y=c,\) and \(y=d\).

Short answer

Question: Justify the interpretation of the double integral \(\int_{R} f(x, y) d(x, y)\) as the volume of the region enclosed between the surface \(z=f(x, y)\) and the planes \(z=0, x=a, x=b, y=c,\) and \(y=d\). Answer: The interpretation of the double integral \(\int_{R} f(x, y) d(x, y)\) as the volume of the region enclosed between the surface \(z=f(x, y)\) and the planes \(z=0, x=a, x=b, y=c,\) and \(y=d\) is justified as the double integral represents the sum of the volumes of infinitely many small rectangular boxes that fill up the region under the graph of the function \(f(x, y)\). The connection between the double integral and volume is made through the concept of Riemann sums, where each term in the Riemann sum represents the volume of a rectangular box with height \(z=f(x_{ij}, y_{ij})\) and base area \(\Delta x_i \cdot \Delta y_j\). The exact volume is calculated by taking the limit of these Riemann sums as the number of subdivisions approaches infinity.

Step-by-step solution

Defining the Double Integral

The double integral \(\int_{R} f(x, y) d(x, y)\) is defined as the limit of the Riemann sums over the rectangular region \(R=[a, b] \times[c, d]\), as we choose infinitely many subdivisions of the region. The Riemann sum is given by $$S_n = \sum_{i=1}^{n} \sum_{j=1}^{m} f(x_{ij}, y_{ij}) \cdot \Delta x_i \cdot \Delta y_j$$ where \(x_{i-1} \leq x_{ij} \leq x_i\), \(y_{j-1} \leq y_{ij} \leq y_j\), \(\Delta x_i = x_i - x_{i-1}\), and \(\Delta y_j = y_j - y_{j-1}\).

Connecting Riemann Sums with Volume

We can think of each term \(f(x_{ij}, y_{ij}) \cdot \Delta x_i \cdot \Delta y_j\) in the Riemann sum as the volume of a rectangular box with height \(z=f(x_{ij}, y_{ij})\) and base area \(\Delta x_i \cdot \Delta y_j\). The sum of these volumes (\(S_n\)) approximates the total volume of the region in \(\mathbb{R}^3\). However, to find the exact volume, we must consider the limit as the number of subdivisions approaches infinity, which is the double integral: $$Volume = \int_{R} f(x, y) d(x, y)$$

Interpretation of the Double Integral

Now that we have connected the double integral with the volume, we can interpret the double integral \(\int_{R} f(x, y) d(x, y)\) as the volume of the region enclosed between the surface \(z=f(x, y)\) and the planes \(z=0, x=a, x=b, y=c,\) and \(y=d\), given that it exists. This is because the double integral represents the sum of the volumes of infinitely many small rectangular boxes that fill up the region under the graph of the function \(f(x, y)\).

Key concepts

Riemann Sums

Riemann sums are a fundamental concept for approximating the integral of a function over a certain domain. They act as the building blocks for double integrals.

In the context of double integrals, particularly over a rectangular region, Riemann sums approximate the volume under a surface.
Each term in a Riemann sum represents an estimated volume of a tiny rectangular prism, or box. The box's height is given by evaluating the function at specific points within the subdivisions of the region. The base area is defined by the dimensions of each subrectangle, expressed as \( \Delta x_i \) and \( \Delta y_j \).

Consider dividing the region \( R = [a, b] \times [c, d] \) into smaller rectangles. By summing the calculated volumes of all these tiny boxes, we approximate the volume under the surface \( z = f(x, y) \).

• The more subdivisions you take, the more accurate your approximation becomes.
• The step of letting the subdivisions go to infinity leads you to the exact solution, the double integral.

Volume Calculation

When calculating volume using double integrals, we are essentially summing up the infinite collection of small volumes that accumulate to form a 3D shape.
The region in question is generally defined by specific boundaries, helping to create a confined space for integration.

Understanding how Riemann sums approximate volumes helps us appreciate double integrals. Double integrals take Riemann sums to the limit, providing an exact calculation.
Think of each small rectangular area in the \( xy \)-plane. Multiply these areas by their respective heights (from the function \( f(x, y) \)). As these become infinitely small, the summation morphs into a precise volume measurement.

• The function \( f(x, y) \) must be non-negative within the given region to ensure the volume simply adds up instead of canceling out.
• Ensure the function is bounded to avoid infinite volumes.

Limits of Integration

The limits of integration are crucial for setting boundaries on the region over which you perform a double integral. They define the rectangle on the \( xy \)-plane that encapsulates the area of interest.

For the exercise at hand, the region is bounded by \( x = a, x = b, \) and \( y = c, y = d \). These limits restrict the calculation to a specific range, ensuring the solved volume is between the surface and specific planes.

• Think of the region as a box's base, defined on the \( xy \)-plane.
• The boundaries help prevent overflow and ensure no parts of the region are left uncalculated.
• When drawing the area, it's best visualized as the rectangle \( R = [a, b] \times [c, d] \).

The choice of limits is not arbitrary but reflects where the function \( f(x, y) \) operates, specifically within the confines of \( R \). These integrations enable you to account for precisely defined slices of the entire graph of the function that appear above this plane, integrating efficiently over the defined region.