Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**Mathematical Enclosures for Farm Fields**
Farmer Jones and his wife, Dr. Jones, decide to build a fence in their field to keep the sheep safe. Since Dr. Jones is a mathematician, she suggests building fences described by the equations \( y = 7x^2 \) and \( y = x^2 + 8 \). Farmer Jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife.
**Problem:**
Given the equations of these curves, determine the area of the enclosed region.
**Explanation:**
1. **Equations of the Fences:**
- First fence: \( y = 7x^2 \)
- Second fence: \( y = x^2 + 8 \)
2. **Finding Intersection Points:**
To find the region enclosed by these two curves, begin by finding their points of intersection. Set the equations equal to each other:
\[ 7x^2 = x^2 + 8 \]
3. **Solving the Equation:**
Subtract \( x^2 \) from both sides:
\[ 6x^2 = 8 \]
\[ x^2 = \frac{8}{6} \]
\[ x^2 = \frac{4}{3} \]
\[ x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3} \]
4. **Setting Up the Integral:**
The area between two curves is found using integration:
\[
A = \int_{a}^{b} \left[ f(x) - g(x) \right] \, dx
\]
Here, \( f(x) = x^2 + 8 \) and \( g(x) = 7x^2 \), with limits of integration \( a = -\frac{2\sqrt{3}}{3} \) and \( b = \frac{2\sqrt{3}}{3} \).
5. **Calculating the Area:**
\[
A = \int_{-\frac{2\sqrt{3}}{3}}^{\frac{2\sqrt{3}}{3}} \left[ (x](https://content.bartleby.com/qna-images/question/c9659bf0-1b68-44e7-9939-e8c666150000/5b351135-68c0-4173-aa6a-69584387491e/i5qdwr5_processed.jpeg)
Transcribed Image Text:**Mathematical Enclosures for Farm Fields**
Farmer Jones and his wife, Dr. Jones, decide to build a fence in their field to keep the sheep safe. Since Dr. Jones is a mathematician, she suggests building fences described by the equations \( y = 7x^2 \) and \( y = x^2 + 8 \). Farmer Jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife.
**Problem:**
Given the equations of these curves, determine the area of the enclosed region.
**Explanation:**
1. **Equations of the Fences:**
- First fence: \( y = 7x^2 \)
- Second fence: \( y = x^2 + 8 \)
2. **Finding Intersection Points:**
To find the region enclosed by these two curves, begin by finding their points of intersection. Set the equations equal to each other:
\[ 7x^2 = x^2 + 8 \]
3. **Solving the Equation:**
Subtract \( x^2 \) from both sides:
\[ 6x^2 = 8 \]
\[ x^2 = \frac{8}{6} \]
\[ x^2 = \frac{4}{3} \]
\[ x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3} \]
4. **Setting Up the Integral:**
The area between two curves is found using integration:
\[
A = \int_{a}^{b} \left[ f(x) - g(x) \right] \, dx
\]
Here, \( f(x) = x^2 + 8 \) and \( g(x) = 7x^2 \), with limits of integration \( a = -\frac{2\sqrt{3}}{3} \) and \( b = \frac{2\sqrt{3}}{3} \).
5. **Calculating the Area:**
\[
A = \int_{-\frac{2\sqrt{3}}{3}}^{\frac{2\sqrt{3}}{3}} \left[ (x
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