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 \(R\). \(f_{1}(x, y)=x^{2}+y^{2}, f_{2}(x, y)=-x^{2}-y^{2}\); \(R\) is the square with corners (0,0) and (2,3) .

Short answer

The volume between the surfaces is 52.

Step-by-step solution

Identify the functions

We are given two surface functions here: \(f_{1}(x, y) = x^{2} + y^{2}\) and \(f_{2}(x, y) = -x^{2} - y^{2}\). These represent the top and bottom surfaces, respectively.

Define the region R

The region \(R\) is defined as the square with corners \((0, 0)\) and \((2, 3)\). This determines our limits of integration as \(0 \leq x \leq 2\) and \(0 \leq y \leq 3\).

Set up the double integral

To find the volume between the surfaces, we set up a double integral with the function \(f_{1}(x, y) - f_{2}(x, y)\). This gives us the integral \(\int_{0}^{2} \int_{0}^{3} ((x^{2} + y^{2}) - (-x^{2} - y^{2})) \, dy \, dx\).

Simplify the integrand

Simplifying the integrand, we have \((x^{2} + y^{2}) - (-x^{2} - y^{2}) = 2x^{2} + 2y^{2}\). So, the integral becomes \(\int_{0}^{2} \int_{0}^{3} (2x^{2} + 2y^{2}) \, dy \, dx\).

Evaluate the inner integral

Integrate \(2x^{2} + 2y^{2}\) with respect to \(y\) from 0 to 3: \(\int_{0}^{3} (2x^{2} + 2y^{2}) \, dy = [2x^{2}y + \frac{2}{3}y^{3}]_{0}^{3}\). This evaluates to \(6x^{2} + 18\).

Evaluate the outer integral

Integrate \(6x^{2} + 18\) with respect to \(x\) from 0 to 2: \(\int_{0}^{2} (6x^{2} + 18) \, dx = [2x^{3} + 18x]_{0}^{2}\). This evaluates to \(16 + 36\), which is \(52\).

Key concepts

Volume Between Surfaces

Calculating the volume between two surfaces in mathematics involves determining how much space is enclosed by these surfaces over a given region. When we have two functions representing surfaces, the volume between them can be visualized as the space enclosed if we were "sandwiching" something between a top surface and a bottom surface. This is often used in solving physics and engineering problems.

The process typically involves setting up a double integral where the area of the base region is integrated over the difference between the top and bottom functions. For the surfaces given by functions \(f_{1}(x, y)\) and \(f_{2}(x, y)\), we calculate the difference \(f_{1}(x, y) - f_{2}(x, y)\), which gives us the heights at each point \((x, y)\) in our region. By integrating these heights across the entire region, we can find the total volume enclosed between these surfaces.

This method is efficient and precise, providing an exact mathematical approach to volume calculation as opposed to estimation methods.

Region of Integration

The region of integration, often denoted as \(R\), is the area in the plane over which the integral is being evaluated. In our example, the region \(R\) is defined as the square with corners at \((0, 0)\) and \((2, 3)\). This gives us explicit limits for the integrals, allowing us to properly determine the boundaries over which the functions will be evaluated.

The choice of region impacts the integral by specifying where the volume computations are relevant. It is essential to correctly interpret the given region and align it with the x and y limits so that the entire area is covered during integration. For integrals over non-standard regions, sometimes these regions might not align perfectly with rectangular bounds, requiring additional transformations of coordinates.

Choosing the correct region is crucial for obtaining accurate results, and careful visualization or sketching the region helps to ensure no part is neglected or over-counted.

Definite Integral

A definite integral computes the accumulation of quantities, such as area or volume, over specified limits. In the context of volumes between surfaces, the definite double integral sums the infinitesimal volumes (tiny elements) under discussion. Mathematically, it is represented by limits on the integral sign, defining the range over which the integration occurs.

For the problem above, the definite integral is set up as \(\int_{0}^{2} \int_{0}^{3} (2x^{2} + 2y^{2}) \, dy \, dx\). The inner integral \(\int_{0}^{3}\) respects to \(y\), assesses the contribution of volume in the vertical direction, and the outer integral \(\int_{0}^{2}\) respects to \(x\), accumulates these contributions across the horizontal axis.

Calculating definite integrals involves finding antiderivatives, evaluating these at the boundaries, and subtracting. In computational practice, this results in precise total sums like the solution here, which concludes with a volume of 52 units.

Setting Up Integrals

Setting up integrals correctly is the foundation of solving problems involving volume between surfaces. The process begins by clearly understanding the functions that define the surfaces and the region over which you will calculate the volume.

In this exercise, the first step is computing the integrand by subtracting the bottom function \(f_{2}(x, y)\) from the top function \(f_{1}(x, y)\), resulting in the expression \((x^{2} + y^{2}) - (-x^{2} - y^{2}) = 2x^{2} + 2y^{2}\). This integrand represents the differential volume elements you intend to sum.

Next, it involves determining the correct limits of integration for both variables \(x\) and \(y\), which are dictated by the dimensions of the region \(R\). Setting these correctly avoids any computational errors. Finally, verify and simplify the integrand if possible before evaluating it, ensuring that the integral is ready for calculation and produces a tangible solution.