Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid whose volume is given by the following double integrals over the rectangle \(R=\\{(x, y)\) : \(0 \leq x \leq 2,0 \leq y \leq 3\\}\). \(\iint_{R}(y+1) d A\)

Step-by-Step Solution

Answer

Sketch a slanted slab over the rectangle from \(z=1\) to \(z=4\).

Step 1: Understand the Problem

We need to sketch the solid whose volume is represented by the double integral \(\iint_{R}(y+1)\,dA\), where the region \(R\) is defined as \((x, y) : 0 \leq x \leq 2, 0 \leq y \leq 3\). This integral represents the volume under the surface \(z = y + 1\) and above the rectangle \(R\) in the \(xy\)-plane.

Step 2: Analyze the Function

The function \(f(x, y) = y + 1\) describes a plane. For any fixed \(y\), the plane has a constant height \(z = y + 1\). As \(y\) changes from 0 to 3, the height of the plane ranges from 1 to 4. This results in a slanted surface.

Step 3: Determine Bounds for the Integral

The integral is taken over the rectangle \(R\), with \(x\) ranging from 0 to 2 and \(y\) ranging from 0 to 3. These are the bounds for our integration.

Step 4: Visualize the Solid

The solid is between the \(xy\)-plane and the plane \(z = y + 1\), over the rectangular region \(0 \leq x \leq 2\) and \(0 \leq y \leq 3\). This means the solid extends vertically from \(z = 0\) up to \(z = 4\) (when \(y = 3\)). You can picture this as a slanted slab, with varying height based on \(y\).

Step 5: Sketch the Solid

To sketch, draw a rectangle on the \(xy\)-plane with vertices at \((0,0), (2,0), (2,3), (0,3)\). Above each point \((x, y)\) on this rectangle, plot it up to the height \(z = y + 1\). Connect these to form a slanting surface from \(z=1\) where \(y=0\) to \(z=4\) where \(y=3\), over the defined rectangle.

Key Concepts

Volume Under a Surface

Understanding the concept of the volume under a surface is key to solving problems involving double integrals. When you have a function, like the one in the exercise, \(z = y + 1\), it describes a surface in three-dimensional space. The double integral \( \iint_{R}(y+1)\,dA \) calculates the volume beneath this surface and above a specified region on the \(xy\)-plane.

In this exercise, the function \(z = y + 1\) represents a plane that is slanted because the height \(z\) increases as \(y\) increases. To find the volume under this surface across a defined region, you sum up all the tiny volumes formed by thin columns standing on the \(xy\)-plane. Each column reaches up to the surface itself, which, in this exercise, is represented by \(z = y + 1\).

The double integral involves a process of lifting an infinitesimally small area dA, which in rectangular coordinates, is \(dA = dx \, dy\) or \(dy \, dx\), to the height determined by \(z = y + 1\). Integrating over this small area computes the volume of each small column, and then summing these over the entire region \(R\) gives the total volume under the surface.

Rectangular Region

The rectangular region \(R\) in double integrals specifies the area on the \(xy\)-plane over which we are calculating the volume under a surface. In our problem, this region \(R = \{(x, y)\) : \(0 \leq x \leq 2, 0 \leq y \leq 3 \}\) is a straightforward rectangle.

This rectangle is defined by the intervals for \(x\) and \(y\):

• \(x\) ranges from 0 to 2
• \(y\) ranges from 0 to 3

These intervals specify the bounds for the double integral. It's like drawing a box on the 2D plane where the calculations will happen. To picture it, imagine drawing a standard rectangle by plotting its corner points in the \(xy\)-plane—these points being \((0,0), (2,0), (2,3), (0,3)\).

Within this rectangular region, every point \((x, y)\) serves as a base for a column of volume that extends upwards to meet the surface \(z = y + 1\). The entire volume under the surface is considered only over this region, and no portion outside this rectangle is included.

Surface Sketching

Sketching the surface in a problem like this can greatly help in visualizing the concept of volume under a surface when working with double integrals. Start with the defined region \(R\) on the \(xy\)-plane, which we've identified as a rectangle from the coordinates provided.

Once you have your base region sketched, consider how the surface \(z = y + 1\) looks above this rectangle. For any point within the rectangle, the height is determined solely by the \(y\)-value, with \(z\) increasing linearly from 1 when \(y = 0\) to 4 when \(y = 3\). This creates a tilted plane or slanting surface over the rectangle.

• To sketch this, draw a rectangle on your paper to represent the base on the \(xy\)-plane.
• Then, at each corner of this rectangle, plot a point where \(z = y + 1\).
• Connect these points to form a surface that raises diagonally.

Visualizing the slanted surface above the rectangle can be challenging but is an important part of understanding double integrals. Remember, the volume you calculate is essentially the space between this surface and the \(xy\)-plane over the specified rectangular region.