Problem 67 - University Calculus: Early Transcendentals

Question

Sketch the region of integration and the solid whose volume is given by the double integral. $$\int_{0}^{3} \int_{0}^{2-2 x / 3}\left(1-\frac{1}{3} x-\frac{1}{2} y\right) d y d x$$

Step-by-Step Solution

Answer

The region is a triangle with vertices at (0,0), (3,0), (0,2), and the solid volume decreases as x and y increase.

Step 1: Understand the Limits of Integration

First, observe the limits of the integral: - The outer integral with respect to "dx" goes from 0 to 3.- The inner integral with respect to "dy" goes from 0 to \(2 - \frac{2}{3}x\).These describe a region in the xy-plane.

Step 2: Define the Region of Integration

The region of integration is bounded by:- The line \(y = 0\) (the x-axis),- The line \(x = 0\) (the y-axis), and- The line \(y = 2 - \frac{2}{3}x\).Additionally, x ranges from 0 to 3.

Step 3: Sketch the Boundary Lines

Plot the lines on the xy-plane:- \ y = 0 \ (horizontal axis)- \ x = 0 \ (vertical axis)- Rewrite \(y = 2 - \frac{2}{3}x\) in slope-intercept form, \(x = \frac{3}{2}(2-y)\), where it intersects the x-axis at x = 3 and the y-axis at y = 2.Draw this line.

Step 4: Shade the Region of Integration

Shade the region of integration:Where \ x \ ranges from 0 to 3 and \ y \ is between the x-axis and the line \(y = 2 - \frac{2}{3}x\). This forms a triangular region with vertices at (0,0), (3,0), and (0,2).

Step 5: Understand the Function of Integration

The function \(1 - \frac{1}{3}x - \frac{1}{2}y\) describes the height of the solid at any point (x, y) within the region. The volume under this function, over the triangular region, is calculated by the double integral.

Step 6: Visualize the Solid

The solid is formed above the triangular region in the xy-plane and below the surface \(z = 1 - \frac{1}{3}x - \frac{1}{2}y\). It is a "slanted" volume decreasing in height as x and y increase within the region.

Key Concepts

Region of Integration

In the context of double integrals, the region of integration plays a crucial role. It defines the area over which the integration is performed. Imagine this as the base of our three-dimensional solid.
First, let's consider the boundaries provided in the problem:

• The line \(y = 0\) acts as the lower boundary, essentially representing the x-axis.
• The line \(x = 0\) is the left vertical boundary, or the y-axis.
• The line \(y = 2 - \frac{2}{3}x\) is slanted, forming a diagonal boundary in the xy-plane. This is where all the action happens! We can rewrite this in the slope-intercept form \(x = \frac{3}{2}(2-y)\), which might make it easier to understand where it intersects the axes: at (3,0) on the x-axis and (0,2) on the y-axis.
• With \(x\) ranging from 0 to 3, the region forms a triangular shape with vertices at (0,0), (3,0), and (0,2).

Inside this triangle, is where the integration occurs, as this defines our region of integration in the xy-plane.

Volume Calculation

The double integral approach gives us a powerful tool to calculate volumes above a region in the xy-plane. In this exercise, we use the double integral to find the volume of a solid above the triangular region we've defined.Let's break it down:

• The function \(1 - \frac{1}{3}x - \frac{1}{2}y\) represents the height of the solid at any given point \((x, y)\).
• Think of this function as a surface hovering above the triangular region we previously discussed.
• As \(x\) and \(y\) increase, this surface gently slopes downwards, making the volume a bit "slanted" rather than a neat box.

Computing the double integral essentially sums up all these tiny slivers of area topped by the height, giving you the total volume. The calculated integral does not only find area but also accounts for varying heights over the region.

XY-Plane

The xy-plane is a critical reference frame in mathematics and physics for visualizing two-dimensional regions. In this scenario, it lays the groundwork for our 3D solid's base.Think of the xy-plane as a flat sheet where you can draw shapes — like the triangle in our exercise — that define regions of interest. Once you have the region (in this case, the triangular region), you can lift a surface above it to create a volume.

• The boundaries in the exercise are all plotted on this plane.
• The region being analyzed lies strictly within the xy-plane.
• Any point \((x, y)\) in the plane corresponds to a location from which we measure upwards to determine volume.

Following these principles allows us to extend our calculations out of just flat surfaces and into tangible shapes and volumes, bridging our understanding from 2D to 3D.