The integral gives the area of the region in the xy-plane. Sketch the region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. 126y 0 1² 120 dx dy
The integral gives the area of the region in the xy-plane. Sketch the region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. 126y 0 1² 120 dx dy
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Calculating the Area of a Region in the xy-plane Using a Double Integral
In this exercise, we are given a double integral that represents the area of a region in the xy-plane. We will follow several steps:
1. **Sketch the region** described by the integral.
2. **Label each bounding curve** with its equation.
3. **Determine the coordinates** of the points where the curves intersect.
4. **Find the area of the region.**
#### Given Integral
\[
\int_{0}^{12} \int_{y^2/2}^{6y} dx \, dy
\]
#### Step-by-Step Explanation
1. **Sketch the Region**
- The limits of integration for \( dy \) are from \( y = 0 \) to \( y = 12 \).
- The limits of integration for \( dx \) are from \( x = y^2/2 \) to \( x = 6y \).
2. **Bounding Curves**
- The first bounding curve is \( x = y^2 / 2 \).
- The second bounding curve is \( x = 6y \).
3. **Intersection Points**
- To find the points where the curves intersect, set \( y^2 / 2 = 6y \).
- Solving the equation \( y^2 / 2 = 6y \) involves rewriting it as \( y^2 - 12y = 0 \).
- Factoring gives \( y(y - 12) = 0 \), yielding \( y = 0 \) and \( y = 12 \).
- Next, plug these \( y \)-values back into their respective \( x \)-equations.
- For \( y = 0 \): \( x = 0^2 / 2 = 0 \) and \( x = 6 \times 0 = 0 \).
- For \( y = 12 \): \( x = 12^2 / 2 = 72 \) and \( x = 6 \times 12 = 72 \).
Therefore, the points of intersection are \( (0,0) \) and \( (72,12) \).
4. **Finding the Area**
- Evaluate the integral:
\[
\int_{0}^{12} \int_{y^2/2}^{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1840ebe-b0df-4fe8-9210-d8e3dcfa32cc%2F72f7ec19-1e59-4c57-922c-48c106c47602%2Frdbp0d8_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Area of a Region in the xy-plane Using a Double Integral
In this exercise, we are given a double integral that represents the area of a region in the xy-plane. We will follow several steps:
1. **Sketch the region** described by the integral.
2. **Label each bounding curve** with its equation.
3. **Determine the coordinates** of the points where the curves intersect.
4. **Find the area of the region.**
#### Given Integral
\[
\int_{0}^{12} \int_{y^2/2}^{6y} dx \, dy
\]
#### Step-by-Step Explanation
1. **Sketch the Region**
- The limits of integration for \( dy \) are from \( y = 0 \) to \( y = 12 \).
- The limits of integration for \( dx \) are from \( x = y^2/2 \) to \( x = 6y \).
2. **Bounding Curves**
- The first bounding curve is \( x = y^2 / 2 \).
- The second bounding curve is \( x = 6y \).
3. **Intersection Points**
- To find the points where the curves intersect, set \( y^2 / 2 = 6y \).
- Solving the equation \( y^2 / 2 = 6y \) involves rewriting it as \( y^2 - 12y = 0 \).
- Factoring gives \( y(y - 12) = 0 \), yielding \( y = 0 \) and \( y = 12 \).
- Next, plug these \( y \)-values back into their respective \( x \)-equations.
- For \( y = 0 \): \( x = 0^2 / 2 = 0 \) and \( x = 6 \times 0 = 0 \).
- For \( y = 12 \): \( x = 12^2 / 2 = 72 \) and \( x = 6 \times 12 = 72 \).
Therefore, the points of intersection are \( (0,0) \) and \( (72,12) \).
4. **Finding the Area**
- Evaluate the integral:
\[
\int_{0}^{12} \int_{y^2/2}^{
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