Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid whose volume is the indicated iterated integral. \(\int_{0}^{1} \int_{0}^{2} \frac{x}{2} d x d y\)

Step-by-Step Solution

Answer

The solid is a triangular prism with a rectangular base from (0,0) to (2,1).

Step 1: Understanding the Integral

The given iterated integral is \( \int_{0}^{1} \int_{0}^{2} \frac{x}{2} \, dx \, dy \). This represents the volume under the function \( z = \frac{x}{2} \) over the region specified in the xy-plane that satisfies \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \).

Step 2: Setting Up the Boundaries

The limits of the integrals indicate that the region of integration in the xy-plane is a rectangle. When \( x \) ranges from 0 to 2 and \( y \) ranges from 0 to 1, the region forms a rectangle with one side on the y-axis (0 to 1) and another parallel to the y-axis along \( x = 2 \).

Step 3: Visualizing the 3D Region

The function \( z = \frac{x}{2} \) describes a surface where the height of any point in the xy-region is \( \frac{x}{2} \). For \( x = 2 \), the highest point is at \( z = 1 \), and when \( x = 0 \), \( z = 0 \). This forms a plane that starts at the line \( y = 0 \) for different x-values in the xy-plane and ends at a maximum height of 1 at \( x = 2 \).

Step 4: Sketching the Solid

To sketch the solid, note that it is confined below the plane \( z = \frac{x}{2} \). The solid is a wedge-like volume that rises from the yz-plane (\( x=0 \) where \( z=0 \)) to the plane \( x=2 \) and \( z=1 \) at its maximum height. It forms a triangular prism resting on the rectangular base \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \).

Key Concepts

Visualization of 3D Regions

When working with iterated integrals, it's important to understand the three-dimensional space they represent. In our case, we are looking at a solid defined by the integral \( \int_{0}^{1} \int_{0}^{2} \frac{x}{2} \, dx \, dy \). This suggests a region in 3D formed under a surface. The function \( z = \frac{x}{2} \) represents the surface over which the solid is confined. To better comprehend this, consider:

• The surface \( z = \frac{x}{2} \) is a plane that progressively increases in height as \( x \) increases from 0 to 2.
• Visualize the xy-plane as a sheet on the floor, bounded by \( x = 0 \) to \( x = 2 \) and \( y = 0 \) to \( y = 1 \).
• The height \( z \) varies only with \( x \), creating a sloped surface that peaks at \( z = 1 \) when \( x = 2 \). This looks like a ramp rising along the length of the rectangular base.

Visualizing these elements helps in understanding the shape and structure of the region under consideration. This 3D sketch forms a triangular prism when you view it from different angles.

Volume Calculation

Calculating the volume of the solid under the surface can be achieved using an iterated integral. Given the integral \( \int_{0}^{1} \int_{0}^{2} \frac{x}{2} \, dx \, dy \), the process generally involves:

• Evaluating the inner integral, which in this case is \( \int_{0}^{2} \frac{x}{2} \, dx \). This evaluates the area under the curve \( \frac{x}{2} \) from \( x = 0 \) to \( x = 2 \).
• After integrating with respect to \( x \), substitute the limits to find a definite value.
• Next, evaluate the outer integral \( \int_{0}^{1} \, dy \) using the result from the inner integral. This adds the third dimension by integrating over the \( y \)-range.

This process results in the volume of the solid, accounting for all points in the 3D region under the constraint \( z = \frac{x}{2} \). Calculating the volume is essential in understanding the total space enclosed by a surface.

Rectangular Integration Region

The rectangular integration region is a critical aspect of these types of integrals. It simplifies the determination of bounds for each part of the integral. Here:

• The bounds for \( x \) and \( y \) in the iterated integral clearly define a rectangle on the xy-plane, \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 1 \).
• Understanding this rectangle helps in comprehending the area over which you are integrating. It acts as the base of the 3D geometry.
• Visualizing this rectangle allows you to see how the function \( z = \frac{x}{2} \) interacts with this area to form the entire solid.

When you approach such problems, always break down the xy-boundary into simple, visual terms. This guarantees you're correctly interpreting the integration limits and the resulting 3D formation.