Find the area of region enclosed by =t-t, y= t-t, 0

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
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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[ -
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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