Problem 57 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the volume of the solid under the surface \(z=f(x, y)\) and over the given region \(R\).$$ f(x, y)=\frac{1}{x y} $$ \(R: 1 \leq x \leq 2,1 \leq y \leq 3\)

Step-by-Step Solution

Answer

The volume under the surface and over the region is \ln(3) \ln(2).

Step 1 - Set Up the Double Integral

To find the volume under the surface and above the given region, set up the double integral of the function over the region. The volume is given by \[ V = \int \int_{R} f(x, y) \,dx \,dy \] with \(f(x, y) = \frac{1}{xy} \) and \(R: 1 \leq x \leq 2, 1 \leq y \leq 3 \).

Step 2 - Write the Integral in Terms of Limits

Since the region \(R\) is rectangular from \(x = 1\) to \(x = 2\) and \(y = 1\) to \(y = 3\), the double integral can be written as: \[ V = \int_{1}^{2} \int_{1}^{3} \frac{1}{xy} \,dy \,dx \]

Step 3 - Integrate with Respect to y

First, integrate with respect to \(y\): \[ \int_{1}^{3} \frac{1}{xy} \,dy = \frac{1}{x} \int_{1}^{3} \frac{1}{y} \,dy \]. The antiderivative of \( \frac{1}{y} \) is \( \ln(y) \), so after integrating, \[ \frac{1}{x} [\ln(y)]_{1}^{3} = \frac{1}{x} (\ln(3) - \ln(1)) \]. \( \ln(1) = 0 \), so this simplifies to \[ \frac{1}{x} \ln(3) \]

Step 4 - Integrate with Respect to x

Now integrate \( \frac{\ln(3)}{x} \) with respect to \(x\): \[ \ln(3) \int_{1}^{2} \frac{1}{x} \,dx \]. The antiderivative of \( \frac{1}{x} \) is \( \ln(x) \), so \[ \ln(3) [\ln(x)]_{1}^{2} = \ln(3) (\ln(2) - \ln(1)) \], and \( \ln(1) = 0 \), so this simplifies to \[ \ln(3) \ln(2) \].

Step 5 - Combine the Results

The final volume is given by \[ V = \ln(3) \ln(2) \].

Key Concepts

double integral

In mathematics, a double integral is used to compute the volume under a surface over a given region. For functions of two variables, this integral helps find the aggregate effect of small elements over a specified area. In this particular exercise, we used a double integral to find the volume under the surface described by the function \(f(x, y) = \frac{1}{xy}\) and over a given rectangular region \(R\). This is expressed as:
\[ V = \int \int_{R} f(x, y) \,dx \,dy \]
To set up a double integral, follow these steps:

• Define the function \(f(x, y)\). In this example, it’s \(\frac{1}{xy}\)
• Identify the limits of integration based on the region \(R\). Here, \(R\) spans from \(x = 1\) to \(x = 2\) and \(y = 1\) to \(y = 3\)

After defining the integral, we can integrate step by step, first with respect to one variable and then the other, to obtain the total volume.

antiderivative

An antiderivative is a function whose derivative is the given function. When we integrate, we look for the antiderivative to determine the accumulated area or volume under a curve or surface. In this problem, we compute two separate antiderivatives: one for \(\frac{1}{y}\) and one for \(\frac{1}{x}\).
Firstly, when integrating with respect to \(y\) for the inner integral:
\[ \frac{1}{x} \, \int_{1}^{3} \, \frac{1}{y} \,dy = \, \frac{1}{x} (\frac{\ln(y)}{1 \, \text{to} \, 3}) = \, \frac{1}{x} (\ln(3) - \ln(1)) \]
Since \(\ln(1) = 0\), this simplifies to \(\frac{1}{x} \ln(3)\)
Next, when integrating with respect to \(x\) for the outer integral:
\[ \ln(3) \, \int_{1}^{2} \, \frac{1}{x} \,dx = \, \ln(3) (\ln(x) \text{from} \ 1 \, \text{to} \, 2) = \, \ln(3) (\ln(2) - \ln(1)) \]
Again, since \(\ln(1) = 0\), the result is \(\ln(3) \ln(2)\)

region R

The region \(R\) is the area over which we are integrating. For double integrals, this region can be described by boundaries for the variables involved. In this exercise, \(R\) is a rectangular area defined by:

• For \(x\), the limits are 1 to 2
• For \(y\), the limits are 1 to 3

With these boundaries, we properly set up our double integral.
\[ V = \int_{1}^{2} \int_{1}^{3} \frac{1}{xy} \,dy \,dx \]
It is crucial to respect the specified limits, as they define the region over which the volume under the surface \(z\) is calculated. By integrating within these limits, we ensure we’re computing the volume within the desired area.