Understanding the Vector Field
The vector field given is:
\[ \vec{F} = y \hat{y} + (x z^3 - y z) \hat{v} \]
In this case, we define \( z = 2 \) for the circular path \( C \).
Substituting the Value of z
Substituting \( z = 2 \) into the vector field:
\[ \vec{F} = y \hat{y} + (x (2^3) - y (2)) \hat{v} \]
This simplifies to:
\[ \vec{F} = y \hat{y} + (8x - 2y) \hat{v} \]
Parametrizing the Circle
The circle \( x^2 + y^2 = 4 \) can be parametrized as:
- \( x = 2 \cos(t) \)
- \( y = 2 \sin(t) \)
- Where \( t \) ranges from \( 0 \) to \( 2\pi \).
Calculating the Line Integral
To find the circulation of the vector field around the curve \( C \), compute:
\[ \oint_C \vec{F} \cdot d\vec{r} \]
where \( d\vec{r} = (dx, dy, dz) \).
Since \( z \) is constant, \( dz = 0 \), and:
\[ d\vec{r} = (-2 \sin(t) dt, 2 \cos(t) dt, 0) \]
Dot Product Calculation
Now, we compute \( \vec{F} \cdot d\vec{r} \):
1. Substitute \( x \) and \( y \) into \( \vec{F} \):
- \( \vec{F} = 2 \sin(t) \hat{y} + (8(2 \cos(t)) - 2(2 \sin(t))) \hat{v} \)
- \( \vec{F} = 2 \sin(t) \hat{y} + (16 \cos(t) - 4 \sin(t)) \hat{v} \)
2. Perform the dot product:
\[ \vec{F} \cdot d\vec{r} = (2 \sin(t) \hat{y} + (16 \cos(t) - 4 \sin(t)) \hat{v}) \cdot (-2 \sin(t) dt, 2 \cos(t) dt, 0) \]
This simplifies to:
\[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} (4 \sin^2(t) - (16 \cos(t) - 4 \sin(t)) 2 \sin(t)) dt \]
Final Integration
Integrating \( 4 \sin^2(t) + 32 \cos(t) \sin(t) - 8 \sin^2(t) \) across \( 0 \) to \( 2\pi \):
- Recognize that the integral of \( \sin(t) \) and \( \cos(t) \) over