Question

Sketch the solid whose volume is given by the following double integrals over the rectangle \(R=\\{(x, y)\) : \(0 \leq x \leq 2,0 \leq y \leq 3\\}\) $$ \iint_{R}(x-y+4) d A $$

Short answer

The solid is a slanted prism over the rectangle defined by \( 0 \leq x \leq 2, 0 \leq y \leq 3 \), with varying height based on \( f(x, y) = x - y + 4 \).

Step-by-step solution

Understand the Region of Integration

We are given the region \( R \) which is defined as \( \{(x, y) : 0 \leq x \leq 2, 0 \leq y \leq 3\} \). This region is a rectangle in the xy-plane with width 2 and height 3. The vertices of this rectangle are \((0, 0), (2, 0), (2, 3), (0, 3)\).

Interpret the Double Integral

The double integral \( \iint_{R}(x-y+4)\, dA \) implies that we are integrating the function \( f(x, y) = x - y + 4 \) over the region \( R \). This function represents the height of the solid above each point \( (x, y) \) in the region \( R \).

Analyze the Function

The function \( f(x, y) = x - y + 4 \) is a plane surface in three dimensions. To understand its shape, consider its intercepts: when \( x = 0 \), \( f(0, y) = 4 - y \); when \( y = 0 \), \( f(x, 0) = x + 4 \); and when \( x = y \), \( f(x, y) = 4 \). This plane is slanted and decreases in the y-direction and increases in the x-direction.

Visualize the Solid

The solid can be visualized as a rectangular prism under the surface \( x - y + 4 \) over the rectangle \( R \). At the corner \( (0, 0) \), the height of the solid is 4. At \( (2, 0) \), it is 6. At \( (0, 3) \), it is 1. At \( (2, 3) \), it is 3. The surface is slanted within these boundary conditions.

Sketch the Solid

To sketch the solid, first draw the base rectangle in the xy-plane with vertices at \((0, 0), (2, 0), (2, 3), (0, 3)\). From each vertex point, extend upward by the height given by \( f(x, y) \). Connect these endpoints to form the slanted, planar surface. It will resemble a tilted roof above the rectangle, forming a solid prism with non-uniform height.

Key concepts

Region of Integration

Regions of integration are foundational in the study of double integrals. In this exercise, our region of integration \( R \) is set on the xy-plane. It's defined as a rectangle with boundaries spanning from \( x = 0 \) to \( x = 2 \) and from \( y = 0 \) to \( y = 3 \).
The vertices of this rectangle are quite straightforward:

• At \((0, 0)\)
• At \((2, 0)\)
• At \((2, 3)\)
• At \((0, 3)\)

This simple rectangular region provides a base upon which the function \( f(x, y) = x - y + 4 \) is evaluated. As you explore more complex shapes of regions, keep in mind that identifying the boundaries clearly is crucial.
Once the region is defined, the double integral computes the volume of the solid above this rectangle in the xy-plane.

Visualizing Solids

Visualizing solids from double integrals involves understanding how the function behaves over the region of integration. The function \( f(x, y) = x - y + 4 \) describes a plane in three-dimensional space. This plane is not flat like a desk surface; rather, it has inclines because its shape changes with \( x \) and \( y \).

• At the point \((0, 0)\), the height is 4.
• Moving to \((2, 0)\), the height increases to 6, showing a slope upward as \( x \) increases.
• At \((0, 3)\), the height is 1, indicating a decrease as \( y \) increases.
• Finally, at \((2, 3)\), the height is 3.

When mentally mapping the heights onto these points, envision a tilted surface that resembles a slanted roof spanning across the rectangle base. By connecting these points, you draw out the picture of the solid—shaped like a prism with varying height.
This three-dimensional perspective aids in grasping how the integral summation calculates the total volume of the solid.

Three-Dimensional Surfaces

Three-dimensional surfaces are key to understanding the physical interpretation of multivariable calculus. In the context of this exercise, the surface is not just theoretical; it models a shape with varied height over a specific region of integration.
The function \( f(x, y) = x - y + 4 \) in this problem creates a slanted plane. This incline can be better understood by looking at the intercepts:

• On the \( x \)-axis, when \( y = 0 \), it's \( f(x, 0) = x + 4 \).
• On the \( y \)-axis, when \( x = 0 \), it's \( f(0, y) = 4 - y \).
• Where \( x = y \), the function remains flat as \( f(x, y) = 4 \).

These intercepts show how the plane will dip and rise across the xy-plane. When sketched, you can see the variations and the clear non-uniform height across the rectangle's domain. Such interpretations allow us to relate well to realistic surfaces and volumes and predict behavior in physics and engineering applications.
Understanding these principles helps visualize more intricate and complex three-dimensional forms formed by different functions.