Question

Let \[ \mathbf{F} = (3 + 2xy) \hat{i} + (x^2 - 3y^2) \hat{j} \] and let \( \mathbf{L} \) be the curve \[ \mathbf{r}(t) = e^t \sin t \, \hat{i} + e^t \cos t \, \hat{j}, \quad 0 \leq t \leq \pi. \] Then \[ \int_{\mathbf{L}} \mathbf{F} \cdot d\mathbf{r} = \]

• A. \( e^{-3t} + 1 \)
• B. \( e^{-6t} + 2 \)
• C. \( e^{6t} + 2 \)
• D. \( e^{3t} + 1 \)

Hint:

For line integrals, parametrize the curve and compute the dot product with the vector field.

Answer 1

The Correct Option is D

Step 1: Parametrization of \( \mathbf{F} \).
We first compute the vector \( d\mathbf{r} \) for the curve \( \mathbf{r}(t) \): \[ d\mathbf{r} = \frac{d}{dt} \left( e^t \sin t \right) \hat{i} + \frac{d}{dt} \left( e^t \cos t \right) \hat{j}. \] Thus, \[ d\mathbf{r} = (e^t \cos t + e^t \sin t) \hat{i} + (e^t \cos t - e^t \sin t) \hat{j}. \]
Step 2: Compute the dot product.
Now, compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \): \[ \mathbf{F} \cdot d\mathbf{r} = \left( 3 + 2xy \right) (e^t \cos t + e^t \sin t) + \left( x^2 - 3y^2 \right) (e^t \cos t - e^t \sin t). \] Substitute \( x = e^t \sin t \) and \( y = e^t \cos t \) into the above expression and simplify the integral. The result gives the answer as \( e^{3t} + 1 \).
Step 3: Conclusion.
Thus, the correct answer is (D).