Evaluate the line integral ry dz for C : r(t) = (e , et+1)", te'), 0

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Chapter1: Functions And Models
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### Line Integral Evaluation

**Problem Statement:**

Evaluate the line integral \(\int_C x y \, dz\) for the curve \(C\) parameterized by:
\[ \mathbf{r}(t) = \left\langle e^{-t}, e^{(t+1)^2}, te^t \right\rangle, \]
where \(0 \leq t \leq 1\). 

Then, repeat the evaluation for \(-C\) in place of \(C\).

**Integrals to be evaluated:**

\[
\int_C xy \, dz = \boxed{\ }
\]
\[
\int_{-C} xy \, dz = \boxed{\ }
\]
Transcribed Image Text:### Line Integral Evaluation **Problem Statement:** Evaluate the line integral \(\int_C x y \, dz\) for the curve \(C\) parameterized by: \[ \mathbf{r}(t) = \left\langle e^{-t}, e^{(t+1)^2}, te^t \right\rangle, \] where \(0 \leq t \leq 1\). Then, repeat the evaluation for \(-C\) in place of \(C\). **Integrals to be evaluated:** \[ \int_C xy \, dz = \boxed{\ } \] \[ \int_{-C} xy \, dz = \boxed{\ } \]
### Line Integral Evaluation Problem

**Problem Statement:**

Evaluate the line integral:

\[ \int_{C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy \]

for the given parametrized curve \( C \):

\[ r(t) = \langle \sqrt{t}, t^2 \rangle \]

where \( 0 \le t \le 1 \).

Then, repeat the evaluation with \( -C \) in place of \( C \).

### Integral Expressions:

\[ \int_{C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy = \]

\[ \int_{-C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy = \]

### Explanation:

1. **Parametrization of Curve \( C \)**:
   - \( x = \sqrt{t} \)
   - \( y = t^2 \)

   The parametrization describes the curve in terms of the parameter \( t \), which ranges from \( 0 \) to \( 1 \).

2. **Evaluating the Integral**:
   - Substitute the parametrization into the line integral expression.
   - Calculate the differentials \( dx \) and \( dy \).
   - Integrate with respect to the parameter \( t \) over the given interval.

3. **Reversing the Curve (\( -C \))**:
   - Consider the reverse parametrization where \( t \) is replaced by \( -t \).
   - Evaluate the integral again for this reversed curve.

Explore the integrals and compute to understand how changes in the parametrization affect the integral's value. This concept is crucial in vector calculus and understanding path dependence in fields described by vector functions.
Transcribed Image Text:### Line Integral Evaluation Problem **Problem Statement:** Evaluate the line integral: \[ \int_{C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy \] for the given parametrized curve \( C \): \[ r(t) = \langle \sqrt{t}, t^2 \rangle \] where \( 0 \le t \le 1 \). Then, repeat the evaluation with \( -C \) in place of \( C \). ### Integral Expressions: \[ \int_{C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy = \] \[ \int_{-C} x^2 y \, dx + x^4 \sqrt{y^2 + 1} \, dy = \] ### Explanation: 1. **Parametrization of Curve \( C \)**: - \( x = \sqrt{t} \) - \( y = t^2 \) The parametrization describes the curve in terms of the parameter \( t \), which ranges from \( 0 \) to \( 1 \). 2. **Evaluating the Integral**: - Substitute the parametrization into the line integral expression. - Calculate the differentials \( dx \) and \( dy \). - Integrate with respect to the parameter \( t \) over the given interval. 3. **Reversing the Curve (\( -C \))**: - Consider the reverse parametrization where \( t \) is replaced by \( -t \). - Evaluate the integral again for this reversed curve. Explore the integrals and compute to understand how changes in the parametrization affect the integral's value. This concept is crucial in vector calculus and understanding path dependence in fields described by vector functions.
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