Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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![**Problem Statement:**
Find the area of the region enclosed by the parametric equations:
\[ x = t - t^3, \]
\[ y = t - t^2, \]
where \( 0 \leq t \leq 1 \), using Green's Theorem.
**Solution:**
To solve this problem, we can apply Green’s Theorem, which relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \).
Green's Theorem is given by:
\[ \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA. \]
For finding the area, we choose \( L = 0 \) and \( M = x \), so that:
\[ \oint_{C} x \, dy = \iint_{D} dA. \]
Now, to find \( dy \) in terms of \( t \):
\[ y = t - t^2 \Rightarrow dy = (1 - 2t) \, dt. \]
Parameters for \( x \) and \( dy \) using \( t \):
- \( x = t - t^3 \)
- \( dy = (1 - 2t) \, dt \)
Thus, the integral becomes:
\[ \int_{0}^{1} (t - t^3)(1 - 2t) \, dt. \]
Expanding and integrating:
\[ \int_{0}^{1} \left( t - 2t^2 - t^3 + 2t^4 \right) dt. \]
Each term individually integrated:
- \(\int_{0}^{1} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{1} = \frac{1}{2} \)
- \(\int_{0}^{1} -2t^2 \, dt = \left[ -\frac{2t^3}{3} \right]_{0}^{1} = -\frac{2}{3} \)
- \(\int_{0}^{1} -t^3 \, dt = \left[ -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4f75436-93ee-4e58-a0e3-afe0fa0e1ea5%2F3526ba4c-44da-41c6-baf7-8b16cb6abcbb%2Fmxb9vm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area of the region enclosed by the parametric equations:
\[ x = t - t^3, \]
\[ y = t - t^2, \]
where \( 0 \leq t \leq 1 \), using Green's Theorem.
**Solution:**
To solve this problem, we can apply Green’s Theorem, which relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \).
Green's Theorem is given by:
\[ \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA. \]
For finding the area, we choose \( L = 0 \) and \( M = x \), so that:
\[ \oint_{C} x \, dy = \iint_{D} dA. \]
Now, to find \( dy \) in terms of \( t \):
\[ y = t - t^2 \Rightarrow dy = (1 - 2t) \, dt. \]
Parameters for \( x \) and \( dy \) using \( t \):
- \( x = t - t^3 \)
- \( dy = (1 - 2t) \, dt \)
Thus, the integral becomes:
\[ \int_{0}^{1} (t - t^3)(1 - 2t) \, dt. \]
Expanding and integrating:
\[ \int_{0}^{1} \left( t - 2t^2 - t^3 + 2t^4 \right) dt. \]
Each term individually integrated:
- \(\int_{0}^{1} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{1} = \frac{1}{2} \)
- \(\int_{0}^{1} -2t^2 \, dt = \left[ -\frac{2t^3}{3} \right]_{0}^{1} = -\frac{2}{3} \)
- \(\int_{0}^{1} -t^3 \, dt = \left[ -
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