906 Chapter 15 Multiple Integrals 0 R FIGURE 15.13 The region of integration in Example 2. R _x² + y² = 1 R L <-1 (a) 1 X 1 Leaves at y-VI-a Enters at y=1-x x Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen. EXAMPLE 2 Calculate [[ R sin.x dA, where R is the triangle in the xy-plane bounded by the x-axis, the line y = x, and the line x = 1. Solution The region of integration is shown in Figure 15.13. If we integrate first with respect to y and next with respect to x, then because x is held fixed in the first integration, we find -X sinx L'inx dy) dx = f[x dx = ' [, ²x]" d dx = X 0 3-0 - f'sit 0 If we reverse the order of integration and attempt to calculate IS, dx dy. sin x X sin.x dx = -cos (1) + 1 ≈ 0.46. we run into a problem because ((sinx)/x) dx cannot be expressed in terms of elemen- tary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you first choose doesn't work, try the other. Some- times neither order will work, and then we may need to use numerical approximations. Finding Limits of Integration We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works.
906 Chapter 15 Multiple Integrals 0 R FIGURE 15.13 The region of integration in Example 2. R _x² + y² = 1 R L <-1 (a) 1 X 1 Leaves at y-VI-a Enters at y=1-x x Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen. EXAMPLE 2 Calculate [[ R sin.x dA, where R is the triangle in the xy-plane bounded by the x-axis, the line y = x, and the line x = 1. Solution The region of integration is shown in Figure 15.13. If we integrate first with respect to y and next with respect to x, then because x is held fixed in the first integration, we find -X sinx L'inx dy) dx = f[x dx = ' [, ²x]" d dx = X 0 3-0 - f'sit 0 If we reverse the order of integration and attempt to calculate IS, dx dy. sin x X sin.x dx = -cos (1) + 1 ≈ 0.46. we run into a problem because ((sinx)/x) dx cannot be expressed in terms of elemen- tary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you first choose doesn't work, try the other. Some- times neither order will work, and then we may need to use numerical approximations. Finding Limits of Integration We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Question 3: Notice that this is above the previous reading exercise.
Explain why one order of
Hint 1: Read Example 2 (p. 906).
![The image provides information and diagrams on how to find the limits for double integrals using both vertical and horizontal cross-sections.
### Using Vertical Cross-Sections
When evaluating a double integral \(\iint_R f(x, y) \, dA\), integrate first with respect to \(y\) and then with respect to \(x\). Follow these steps:
1. **Sketch**: Draw the region of integration and label the bounding curves (as demonstrated in Figure 15.14a).
2. **Find the \(y\)-limits of integration**: Imagine a vertical line \(L\) cutting through region \(R\) in the direction of increasing \(y\). Mark the \(y\)-values where \(L\) enters and leaves. These are the \(y\)-limits of integration and they are usually functions of \(x\) (instead of constants) (see Figure 15.14b).
3. **Find the \(x\)-limits of integration**: Choose \(x\)-limits that include all the vertical lines through \(R\). The integral shown here (see Figure 15.14c) is:
\[
\int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} f(x, y) \, dy \, dx.
\]
### Using Horizontal Cross-Sections
To evaluate the same double integral as an iterated integral with the reverse order of integration, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is:
\[
\int_{y=0}^{y=1} \int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}} f(x, y) \, dx \, dy.
\]
### Diagram Descriptions
- **Figure 15.14**: This illustration helps visualize finding the limits of integration:
- **Figure 15.14(a)**: Shows the region of integration with labeled curves.
- **Figure 15.14(b)**: Illustrates the vertical line \(L\) and marks where it enters and leaves the region.
- **Figure 15.14(c)**: Demonstrates the component of the integral using vertical cross-sections, with the smallest \(x\) as \(x=0\) and the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd35cf2d-d59c-46d2-9acb-f7bd2544420e%2Fbd0a6a87-6fe4-4844-8695-4246a4a8d12d%2Fz8x44g_processed.png&w=3840&q=75)
Transcribed Image Text:The image provides information and diagrams on how to find the limits for double integrals using both vertical and horizontal cross-sections.
### Using Vertical Cross-Sections
When evaluating a double integral \(\iint_R f(x, y) \, dA\), integrate first with respect to \(y\) and then with respect to \(x\). Follow these steps:
1. **Sketch**: Draw the region of integration and label the bounding curves (as demonstrated in Figure 15.14a).
2. **Find the \(y\)-limits of integration**: Imagine a vertical line \(L\) cutting through region \(R\) in the direction of increasing \(y\). Mark the \(y\)-values where \(L\) enters and leaves. These are the \(y\)-limits of integration and they are usually functions of \(x\) (instead of constants) (see Figure 15.14b).
3. **Find the \(x\)-limits of integration**: Choose \(x\)-limits that include all the vertical lines through \(R\). The integral shown here (see Figure 15.14c) is:
\[
\int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} f(x, y) \, dy \, dx.
\]
### Using Horizontal Cross-Sections
To evaluate the same double integral as an iterated integral with the reverse order of integration, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is:
\[
\int_{y=0}^{y=1} \int_{x=-\sqrt{1-y^2}}^{x=\sqrt{1-y^2}} f(x, y) \, dx \, dy.
\]
### Diagram Descriptions
- **Figure 15.14**: This illustration helps visualize finding the limits of integration:
- **Figure 15.14(a)**: Shows the region of integration with labeled curves.
- **Figure 15.14(b)**: Illustrates the vertical line \(L\) and marks where it enters and leaves the region.
- **Figure 15.14(c)**: Demonstrates the component of the integral using vertical cross-sections, with the smallest \(x\) as \(x=0\) and the
![# Chapter 15: Multiple Integrals
## Example 2
### Calculate
\[
\int \int_R \frac{\sin x}{x} \, dA,
\]
where \( R \) is the triangle in the xy-plane bounded by the x-axis, the line \( y = x \), and the line \( x = 1 \).
### Solution
The region of integration is shown in Figure 15.13.
**Description of Figure 15.13:**
- A graph illustrating the region \( R \), which is a right triangle.
- The triangle is bounded by the line \( y = x \), the x-axis, and the vertical line \( x = 1 \).
To evaluate the integral, we integrate first with respect to \( y \) and then with respect to \( x \), since \( x \) is held fixed in the first integration.
The calculation steps are:
\[
\int_0^1 \left( \int_0^x \frac{\sin x}{x} \, dy \right) dx = \int_0^1 \left[ \frac{\sin x}{x} \cdot y \right]_0^x dx = \int_0^1 \sin x \, dx = -\cos(1) + 1 \approx 0.46.
\]
If we reverse the order of integration and attempt to calculate
\[
\int_0^1 \left( \int_y^1 \frac{\sin x}{x} \, dx \right) dy,
\]
we encounter a problem because \( \int \frac{\sin(x)}{x} \, dx \) cannot be expressed in terms of elementary functions (there is no simple antiderivative).
### Note
There is no general rule for predicting which order of integration will be the best in situations like this. If the order you first choose doesn’t work, try the other. Sometimes neither order will work, and numerical methods may be needed.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd35cf2d-d59c-46d2-9acb-f7bd2544420e%2Fbd0a6a87-6fe4-4844-8695-4246a4a8d12d%2Fhro3zao_processed.png&w=3840&q=75)
Transcribed Image Text:# Chapter 15: Multiple Integrals
## Example 2
### Calculate
\[
\int \int_R \frac{\sin x}{x} \, dA,
\]
where \( R \) is the triangle in the xy-plane bounded by the x-axis, the line \( y = x \), and the line \( x = 1 \).
### Solution
The region of integration is shown in Figure 15.13.
**Description of Figure 15.13:**
- A graph illustrating the region \( R \), which is a right triangle.
- The triangle is bounded by the line \( y = x \), the x-axis, and the vertical line \( x = 1 \).
To evaluate the integral, we integrate first with respect to \( y \) and then with respect to \( x \), since \( x \) is held fixed in the first integration.
The calculation steps are:
\[
\int_0^1 \left( \int_0^x \frac{\sin x}{x} \, dy \right) dx = \int_0^1 \left[ \frac{\sin x}{x} \cdot y \right]_0^x dx = \int_0^1 \sin x \, dx = -\cos(1) + 1 \approx 0.46.
\]
If we reverse the order of integration and attempt to calculate
\[
\int_0^1 \left( \int_y^1 \frac{\sin x}{x} \, dx \right) dy,
\]
we encounter a problem because \( \int \frac{\sin(x)}{x} \, dx \) cannot be expressed in terms of elementary functions (there is no simple antiderivative).
### Note
There is no general rule for predicting which order of integration will be the best in situations like this. If the order you first choose doesn’t work, try the other. Sometimes neither order will work, and numerical methods may be needed.
---
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 1 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
