Problem 43 - Calculus an Applied Approach

Question

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ 2 x-3 y=0, x+y=5, y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the given equations is \(\frac{25}{6}\).

Step 1: Plot the function

Plot each function (i.e., \(2x-3y=0\), \(x+y=5\), and \(y=0\)) in the xy-plane to visualize the region for which the area needs to be found.

Step 2: Determine the intersecting points

Solve for the intersection points of \(2x-3y=0\), \(x+y=5\), and \(y=0\). As \(2x-3y=0\) can be rewritten as \(y = \frac{2}{3}x\), and setting \(y=0\), we find the x-intersects for this line are \(x=0\) and \(x=5\). We can see these are already the intersection points of the other two functions.

Step 3: Set up the double integral

The general form of the double integral used is \(\int_a^b \int_{f1(x)}^{f2(x)} dy dx\). It is important to decide correctly which function should go below and above in the integral, based on the graphs. Here we have \(f1(x)=0\) and \(f2(x)=\frac{2}{3}x\) when \(x\) is between 0 and 2.5, while \(f1(x)=0\) and \(f2(x)=5-x\) when \(x\) is between 2.5 and 5.

Step 4: Evaluate the integral

Now we evaluate the double integral to find the area. This gives \(\int_0^{2.5} \int_{0}^{\frac{2}{3}x} dy dx + \int_{2.5}^{5} \int_{0}^{5-x} dy dx = \frac{25}{6}\).

Key Concepts

Double Integral

A double integral is a way to calculate volume under a surface over a region in two-dimensional space. It's often used for finding areas of regions bounded by curves in the xy-plane. To evaluate a double integral, we integrate a function of two variables, say \( f(x, y) \), across a specified region in the plane.

The general form of a double integral for area calculation is:

• \( \int\int_R 1 \, dA \), where \( R \) is the region of integration.

In our problem, the double integral is set up by determining the bounds based on the intersection of graphs. First, we identify limits on \( x \) and then determine the corresponding limits on \( y \), given by the functions that form these boundaries. By splitting the calculation into regions where different functions act as the boundary, we find the area of these different regions.

Area Calculation

Area calculation using double integrals involves understanding the region we're integrating over. The double integral

\( \int_a^b \int_{f_1(x)}^{f_2(x)} dy \, dx \)

allows one to calculate the area between two functions, \( f_1(x) \) and \( f_2(x) \), between \( x = a \) and \( x = b \). This structure is significant because it automatically accounts for the geometry of the region based on the functions provided.

In this specific example, the problem asked to calculate the area of a region bounded by:

• \( 2x - 3y = 0 \)
• \( x + y = 5 \)
• \( y = 0 \)

By performing the integral in two parts—one for each subsection of the region defined by different linear functions—it crucially adds flexibility to handle the changes in boundaries within the integration. Integrating over multiple subregions ensures that the calculation thoroughly covers the entire area in question.

Intersection Points

Finding intersection points of functions informs how the regions are bounded. These points are crucial in identifying the limits for our integrals. To locate intersection points algebraically, we solve for where the functions intersect, which usually involves setting them equal and solving the resulting equations.

For the functions in the problem:

• Set \( 2x - 3y = 0 \) equal to \( x + y = 5 \): By substituting \( y = \frac{2}{3}x \) into \( x + y = 5 \), one can find one intersection at \((3, 2)\).
• The lines \( y = \frac{2}{3}x \) and \( y = 0 \) intersect at \( x = 0 \).
• Lastly, the line \( x + y = 5 \) meets the x-axis where \( x = 5 \), \( y = 0 \).

Knowing these points, the limits for \( x \) and the corresponding \( y \) values are determined, helping correctly set up the bounds of integration for our problem.

Graphing Functions

Graphing functions provides a visual representation of the problem, which is especially useful in understanding how regions are defined and where they intersect. By plotting each function on the xy-plane:

• \( 2x - 3y = 0 \) is rewritten as \( y = \frac{2}{3}x \), a line through the origin with a slope of \( \frac{2}{3} \).
• \( x + y = 5 \) is rearranged to \( y = 5 - x \), a line intersecting the y-axis at \( y = 5 \) and the x-axis at \( x = 5 \).
• The line \( y = 0 \) is simply the x-axis.

These plots reveal not only the shape and size of the integrable region but also help verify the accuracy of algebraic calculations for intersections. The visuals guide our understanding of which segments form the boundaries of the enclosed area, prompting an accurate setup of the double integral.