Book cover for Calculus: Early Transcendentals

Calculus: Early Transcendentals

James Stewart

ISBN #9781285741550

8th Edition

6,422 Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section introduces double integrals, extending the concept of the definite integral to functions of two variables to compute areas, volumes, and overall accumulated quantities over a region. By partitioning a region into subrectangles and summing the corresponding volumes, we form double Riemann sums whose limit defines the double integral. Fubini’s Theorem allows for evaluating these integrals as iterated integrals, and numerical methods like the Midpoint Rule can be applied for approximations. The concepts also relate to finding the average value of a function over a region, making them essential in both theoretical and applied contexts.

Learning Objectives

1

Describe the concept of the double integral and its definition via Riemann sums.

2

Explain how double integrals are used to calculate volumes under surfaces over a rectangular region.

3

Demonstrate the construction of iterated integrals and apply Fubini's Theorem to evaluate them.

4

Apply the Midpoint Rule and other numerical methods to approximate double integrals.

5

Interpret the physical meaning of double integrals in contexts such as average value and volume estimation.

Key Concepts

CONCEPT

DEFINITION

Double Integral

An extension of the definite integral to functions of two variables, defined as the limit of double Riemann sums over a region R. It represents quantities such as volume under a surface.

Double Riemann Sum

A sum that approximates the volume under a surface by dividing a region R into subrectangles and summing the product of the function value at a chosen point in each subrectangle and its area.

Iterated Integral

The evaluation of a double integral by performing integration sequentially with respect to one variable and then the other. Fubini’s Theorem guarantees that for continuous functions, the order of integration can be interchanged.

Fubini’s Theorem

A theorem which states that if a function is continuous (or sufficiently well-behaved) on a rectangular region, then the double integral can be computed as an iterated integral in either order.

Midpoint Rule (for double integrals)

A numerical method for estimating the value of a double integral by evaluating the function at the center (midpoint) of each subrectangle and summing the products of these values with the area of the subrectangles.

Average Value of a Function (over a region)

Defined as the total integral of the function over a region divided by the area of the region; it represents the uniform value whose total 'volume' equals the actual accumulated quantity under the function’s surface.

Example Problems

Example 1

(a) Estimate the volume of the solid that lies below the surface $ z = xy $ and above the rectangle $$ R = \{(x, y) \mid 0 \le x \le 6, 0 \le y \le 4 \} $$ Use a Riemann sum with $ m = 3 $, $ n = 2 $, and take the sample point to be the upper right corner of each square. (b) Use the Midpoint Rule to estimate the volume of the solid in part (a).

Example 2

If $ R = [0, 4] \times [-1, 2] $, use a Riemann sum with $ m = 2 $, $ n = 3 $ to estimate the value of $ \iint_R (1 - xy^2)\ dA $. Take the sample points to be (a) the lower right corners and (b) the upper left corners of the rectangles.

Example 3

(a) Use a Riemann sum with $ m = n = 2 $ to estimate the value of $ \iint_R xe^{-xy}\ dA $, where $ R = [0, 2] \times [0, 1] $. Take the sample points to be upper right corners. (b) Use the Midpoint Rule to estimate the integral in part (a).

Example 4

(a) Estimate the volume of the solid that lies below the surface $ z = 1 + x^2 + 3y $ and above the rectangle $ R = [1, 2] \times [0, 3] $. Use a Riemann sum with $ m = n = 2 $ and choose the sample points to be lower left corners. (b) Use the Midpoint Rule to estimate the volume in part (a).

Example 5

Let $ V $ be the volume of the solid that lies under the graph of $ f(x, y) = \sqrt{52 - x^2 - y^2} $ and above the rectangle given by $ 2 \le x \le 4, 2 \le y \le 6 $. Use the lines $ x = 3 $ and $ y = 4 $ to divide $ R $ into subrectangles. Let $ L $ and $ U $ be the Riemann sums computed using lower left corners and upper right corners, respectively. Without calculating the numbers $ V $, $ L $, and $ U $, arrange them in increasing order and explain your reasoning.
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Step-by-Step Explanations

QUESTION

How can we approximate the volume under a surface f(x, y) over a rectangular region R using a double Riemann sum?

STEP-BY-STEP ANSWER:

Step 1: Divide the rectangle R into m subintervals along x and n subintervals along y, forming subrectangles with area ΔA = Δx Δy.
Step 2: Choose a sample point (x*ij, y*ij) in each subrectangle Rij. A common choice is the upper right-hand corner or the center.
Step 3: Compute the approximate volume as the sum: V ≈ Σ (over i=1 to m and j=1 to n) f(x*ij, y*ij) ΔA.
Step 4: Take the limit as m, n → ∞ to define the double integral, which gives the exact volume under the surface.
Final Answer:

Volume Estimation Using Double Integrals

QUESTION

How can a double integral be evaluated as an iterated integral using Fubini's Theorem?

STEP-BY-STEP ANSWER:

Step 1: Write the double integral over a rectangular region R as ∬_R f(x, y) dA.
Step 2: Express it as an iterated integral by fixing one variable; for example, integrate with respect to y first: ∫_(x=a)^(b) [∫_(y=c)^(d) f(x, y) dy] dx.
Step 3: Evaluate the inner integral with respect to y, treating x as a constant, to obtain a function A(x).
Step 4: Integrate the resulting function A(x) with respect to x from a to b.
Step 5: According to Fubini’s Theorem, if f is continuous on R, the order of integration can be switched, providing flexibility in computation.
Final Answer:

Iterated Integral and Fubini’s Theorem

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Common Mistakes

  • Choosing sample points improperly in the Riemann sum, which can lead to inaccurate approximations.
  • Confusing the physical interpretation of the double integral for positive functions (volume) with the average value, especially when functions take negative values.
  • Ignoring the conditions required for Fubini's Theorem, leading to incorrect assumptions about switching the order of integration.
  • Overlooking the need for the dimensions of all subrectangles to approach zero in the limiting process.
  • Misinterpreting the bounds of integration when setting up the iterated integral, which can lead to errors in the final result.