Problem 54 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Use a double integral to find the area of \(R\).\(R\) is the region bounded by \(y=\frac{16}{x}, y=x\), and \(x=8\).

Step-by-Step Solution

Answer

The area of region R is \( 24 - 16 \ln 2 \).

Step 1: Determine the bounds of integration

Identify the region of integration. The curves are given as: 1. \( y = \frac{16}{x} \)2. \( y = x \) The region is bounded between these curves and \( x = 8 \). To find the intersection points, set \( y = \frac{16}{x} \) equal to \( y = x \):\[ x = \frac{16}{x} \]Solving for \( x \): \[ x^2 = 16 \] \[ x = 4 \] (since \( x \) must be positive for this region)

Step 2: Set up the double integral

Set up the double integral to find the area of the region R. The region is between \( x = 4 \) and \( x = 8 \) with the upper curve being \( y = x \) and the lower curve being \( y = \frac{16}{x} \). Thus, the double integral is:\[ \iint_R dA = \int_{4}^{8} \int_{\frac{16}{x}}^{x} dy \, dx \]

Step 3: Evaluate the inner integral

Evaluate the inner integral with respect to \( y \):\[ \int_{\frac{16}{x}}^{x} dy = y \bigg|_{\frac{16}{x}}^{x} = x - \frac{16}{x} \].The result of the inner integral is: \[ \int_{4}^{8} \left(x - \frac{16}{x} \right) dx \]

Step 4: Evaluate the outer integral

Evaluate the outer integral with respect to \( x \):\[ \int_{4}^{8} \left(x - \frac{16}{x} \right) dx = \int_{4}^{8} x \, dx - \int_{4}^{8} \frac{16}{x} \, dx \]Calculate each part separately:\[ \int_{4}^{8} x \, dx = \frac{x^2}{2} \bigg|_{4}^{8} = \frac{64}{2} - \frac{16}{2} = 24 \]\[ \int_{4}^{8} \frac{16}{x} \, dx = 16 \ln|x| \bigg|_{4}^{8} = 16 \ln(8) - 16 \ln(4) \]Simplify the logarithmic terms:\[ 16 (\ln(8) - \ln(4)) = 16 \ln\left(\frac{8}{4}\right) = 16 \ln(2) \]Hence:\[ 16 \cdot \ln(2) \]Combine results:\[ 24 - 16 \ln(2) \]

Key Concepts

Integration Bounds

The first critical step in solving a double integral problem is determining the bounds of integration. To find the area of a given region, we need to know where it starts and ends. In the provided exercise, we're looking at the region bounded by the curves:

• \(y = \frac{16}{x}\)
• y = x

and the vertical line \(x = 8\). To figure out the bounds for the double integral, we need to find the intersection points of the curves. Setting \(y = \frac{16}{x}\) equal to \(y = x\) gives us:\[x = \frac{16}{x}\]Solving this equation, we get \(x^2 = 16\) and thus \(x = 4\) (since \(x\) must be positive).

So, the bounds of integration for \(x\) are from 4 to 8, and for \(y\), it ranges from \(\frac{16}{x}\) (lower curve) to \(x\) (upper curve).

Double Integrals

Double integrals allow us to compute the volume under a surface over a given region. In this problem, they help us find the area of the region \(R\) bounded by specific curves. Once we have our integration bounds, we can set up the double integral as follows:\[\iint_R dA = \int_{4}^{8} \int_{\frac{16}{x}}^{x} dy \, dx\]This integral is set up to first integrate with respect to \(y\), then with respect to \(x\). The inner integral \(\int_{\frac{16}{x}}^{x} dy\) computes the width slice by slice from the lower to the upper bound.

After evaluating the inner integral, we get:\[\int_{\frac{16}{x}}^{x} dy = y \bigg|_{\frac{16}{x}}^{x} = x - \frac{16}{x}\]The problem now reduces to a single integral:\[ \int_{4}^{8} \left(x - \frac{16}{x} \right) dx\]We then break down and solve each part of it separately.

Calculus Applications

Double integrals have various practical applications in calculus, from finding areas and volumes to more complex scenarios such as mass, center of mass, and evaluating moments of inertia. Here, we use them to find the area of a specific region bounded by given curves.
The process is essentially about summing up infinitesimal elements (tiny parts of the whole) to find a total - in this case, the total area. Evaluating each step thoroughly ensures you get the correct final value.

• First, set up the bounds correctly.
• Second, evaluate the inner integral.
• Finally, solve the outer integral.

These steps are not just limited to finding areas but can be adapted for various calculations involving double integrals.
When you've mastered these steps, you can tackle more challenging problems and apply these methods to real-world problems.