Question

Let \( \vec{F}(x, y) = -y \hat{i} + x \hat{j} \) and let \( C \) be the ellipse

\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

with counterclockwise orientation. Then the value of \( \oint_C \vec{F} \cdot d\vec{r} \) (rounded to 2 decimal places) is .............

Hint:

For closed paths, if the vector field is conservative or if the integral is over a full period of a periodic function, the line integral might be zero.

Answer 1

Correct Answer: 75.35

Step 1: Parametrize the ellipse.
The given ellipse can be parametrized using: \[ x = 4 \cos(t), \quad y = 3 \sin(t), \quad \text{where} \quad t \in [0, 2\pi]. \]
Step 2: Find the vector field \( \vec{F}(x, y) \).
We are given the vector field \( \vec{F}(x, y) = -y \hat{i} + x \hat{j} \). Substituting the parametrized values of \( x \) and \( y \), we get: \[ \vec{F}(x, y) = -3 \sin(t) \hat{i} + 4 \cos(t) \hat{j}. \]
Step 3: Find the differential vector \( d\vec{r} \).
The differential displacement vector \( d\vec{r} \) is: \[ d\vec{r} = \frac{d\vec{r}}{dt} dt = (-4 \sin(t) \hat{i} + 3 \cos(t) \hat{j}) dt. \]
Step 4: Compute the dot product \( \vec{F} \cdot d\vec{r} \).
We now compute the dot product \( \vec{F} \cdot d\vec{r} \): \[ \vec{F} \cdot d\vec{r} = (-3 \sin(t) \hat{i} + 4 \cos(t) \hat{j}) \cdot (-4 \sin(t) \hat{i} + 3 \cos(t) \hat{j}). \] This simplifies to: \[ \vec{F} \cdot d\vec{r} = 12 \sin^2(t) - 12 \cos^2(t) = 12(\sin^2(t) - \cos^2(t)). \]
Step 5: Integrate over the interval \( t \in [0, 2\pi] \).
Now, we integrate \( \vec{F} \cdot d\vec{r} \) over one complete revolution of the ellipse: \[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} 12(\sin^2(t) - \cos^2(t)) dt. \] Using the identity \( \sin^2(t) - \cos^2(t) = -\cos(2t) \), we get: \[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} -12 \cos(2t) dt. \] The integral of \( \cos(2t) \) over \( 0 \) to \( 2\pi \) is zero: \[ \oint_C \vec{F} \cdot d\vec{r} = 0. \]
Step 6: Conclusion.
Thus, the value of the line integral is \( \boxed{0} \).