**Sketch the region of integration and evaluate by changing to polar coordinates:** \[ \int_{0}^{1/2} \int_{\sqrt{3x}}^{\sqrt{1-x^2}} 6x \, dy \, dx = \] **Description:** This problem involves a double integral with variable limits of integration, which require sketching the region and converting to polar coordinates. **Integral Details:** 1. The outer integral is with respect to \(x\) from 0 to \(\frac{1}{2}\). 2. The inner integral is with respect to \(y\), ranging from \(\sqrt{3x}\) to \(\sqrt{1-x^2}\). 3. The integrand is \(6x\). **Conversion to Polar Coordinates:** To solve this integral, convert the Cartesian coordinates to polar coordinates \((r, \theta)\), where: - \(x = r \cos \theta\) - \(y = r \sin \theta\) Sketch the region in the \(xy\)-plane defined by: - The lower bound of \(y = \sqrt{3x}\) - The upper bound of \(y = \sqrt{1-x^2}\) These bounds outline a specific area that can be simplified using polar coordinates for easier evaluation of the integral. **Problem Statement:** Calculate the double integral of \( f(x, y) \) over the triangle indicated in the following figure: \( f(x, y) = -14ye^x \) **Explanation of the Figure:** The figure is a graph with \( x \)-axis labeled from 0 to 5 and \( y \)-axis labeled from 0 to 4. It shows a right-angled triangle with vertices at the points (1,1), (5,1), and (1,4). - The hypotenuse of the triangle appears to be inclined downwards from (1,4) to (5,1). **Solution:** The solution to the problem is given as: \[ \text{Answer:} \quad \frac{441}{8} - \frac{189e^4}{8} \]
**Sketch the region of integration and evaluate by changing to polar coordinates:** \[ \int_{0}^{1/2} \int_{\sqrt{3x}}^{\sqrt{1-x^2}} 6x \, dy \, dx = \] **Description:** This problem involves a double integral with variable limits of integration, which require sketching the region and converting to polar coordinates. **Integral Details:** 1. The outer integral is with respect to \(x\) from 0 to \(\frac{1}{2}\). 2. The inner integral is with respect to \(y\), ranging from \(\sqrt{3x}\) to \(\sqrt{1-x^2}\). 3. The integrand is \(6x\). **Conversion to Polar Coordinates:** To solve this integral, convert the Cartesian coordinates to polar coordinates \((r, \theta)\), where: - \(x = r \cos \theta\) - \(y = r \sin \theta\) Sketch the region in the \(xy\)-plane defined by: - The lower bound of \(y = \sqrt{3x}\) - The upper bound of \(y = \sqrt{1-x^2}\) These bounds outline a specific area that can be simplified using polar coordinates for easier evaluation of the integral. **Problem Statement:** Calculate the double integral of \( f(x, y) \) over the triangle indicated in the following figure: \( f(x, y) = -14ye^x \) **Explanation of the Figure:** The figure is a graph with \( x \)-axis labeled from 0 to 5 and \( y \)-axis labeled from 0 to 4. It shows a right-angled triangle with vertices at the points (1,1), (5,1), and (1,4). - The hypotenuse of the triangle appears to be inclined downwards from (1,4) to (5,1). **Solution:** The solution to the problem is given as: \[ \text{Answer:} \quad \frac{441}{8} - \frac{189e^4}{8} \]
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
![**Sketch the region of integration and evaluate by changing to polar coordinates:**
\[
\int_{0}^{1/2} \int_{\sqrt{3x}}^{\sqrt{1-x^2}} 6x \, dy \, dx =
\]
**Description:**
This problem involves a double integral with variable limits of integration, which require sketching the region and converting to polar coordinates.
**Integral Details:**
1. The outer integral is with respect to \(x\) from 0 to \(\frac{1}{2}\).
2. The inner integral is with respect to \(y\), ranging from \(\sqrt{3x}\) to \(\sqrt{1-x^2}\).
3. The integrand is \(6x\).
**Conversion to Polar Coordinates:**
To solve this integral, convert the Cartesian coordinates to polar coordinates \((r, \theta)\), where:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
Sketch the region in the \(xy\)-plane defined by:
- The lower bound of \(y = \sqrt{3x}\)
- The upper bound of \(y = \sqrt{1-x^2}\)
These bounds outline a specific area that can be simplified using polar coordinates for easier evaluation of the integral.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3444a3f-7d5c-4474-ab08-7930ed9e066d%2F2ecefabb-aac5-4558-b6ea-3acdf6df7851%2Fypg5m7m_processed.png&w=3840&q=75)
Transcribed Image Text:**Sketch the region of integration and evaluate by changing to polar coordinates:**
\[
\int_{0}^{1/2} \int_{\sqrt{3x}}^{\sqrt{1-x^2}} 6x \, dy \, dx =
\]
**Description:**
This problem involves a double integral with variable limits of integration, which require sketching the region and converting to polar coordinates.
**Integral Details:**
1. The outer integral is with respect to \(x\) from 0 to \(\frac{1}{2}\).
2. The inner integral is with respect to \(y\), ranging from \(\sqrt{3x}\) to \(\sqrt{1-x^2}\).
3. The integrand is \(6x\).
**Conversion to Polar Coordinates:**
To solve this integral, convert the Cartesian coordinates to polar coordinates \((r, \theta)\), where:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
Sketch the region in the \(xy\)-plane defined by:
- The lower bound of \(y = \sqrt{3x}\)
- The upper bound of \(y = \sqrt{1-x^2}\)
These bounds outline a specific area that can be simplified using polar coordinates for easier evaluation of the integral.
![**Problem Statement:**
Calculate the double integral of \( f(x, y) \) over the triangle indicated in the following figure:
\( f(x, y) = -14ye^x \)
**Explanation of the Figure:**
The figure is a graph with \( x \)-axis labeled from 0 to 5 and \( y \)-axis labeled from 0 to 4. It shows a right-angled triangle with vertices at the points (1,1), (5,1), and (1,4).
- The hypotenuse of the triangle appears to be inclined downwards from (1,4) to (5,1).
**Solution:**
The solution to the problem is given as:
\[ \text{Answer:} \quad \frac{441}{8} - \frac{189e^4}{8} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3444a3f-7d5c-4474-ab08-7930ed9e066d%2F2ecefabb-aac5-4558-b6ea-3acdf6df7851%2Fkujackb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Calculate the double integral of \( f(x, y) \) over the triangle indicated in the following figure:
\( f(x, y) = -14ye^x \)
**Explanation of the Figure:**
The figure is a graph with \( x \)-axis labeled from 0 to 5 and \( y \)-axis labeled from 0 to 4. It shows a right-angled triangle with vertices at the points (1,1), (5,1), and (1,4).
- The hypotenuse of the triangle appears to be inclined downwards from (1,4) to (5,1).
**Solution:**
The solution to the problem is given as:
\[ \text{Answer:} \quad \frac{441}{8} - \frac{189e^4}{8} \]
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