Evaluating the Line Integral

We need to calculate the line integral $\int_C F \cdot dr$ where the vector field is $F(x, y, z) = -y\hat{i}+x\hat{j}+z\hat{k}$ and the curve $C$ is given by $r(t) = \cos t \ \hat{i} + \sin t \ \hat{j} + 2t^2\hat{k}$ for $0 \leq t \leq 1$.

Parametric Representation

The curve $C$ is parameterized by:

• $x(t) = \cos t$
• $y(t) = \sin t$
• $z(t) = 2t^2$

Calculating $dr$

First, find the derivative of $r(t)$ with respect to $t$:

$r'(t) = \frac{dr}{dt} = -\sin t \ \hat{i} + \cos t \ \hat{j} + 4t\hat{k}$

Therefore, $dr = r'(t) dt = (-\sin t \ \hat{i} + \cos t \ \hat{j} + 4t\hat{k}) dt$.

Evaluating $F$ along $C$

Substitute the parametric equations into the vector field $F$:

$F(r(t)) = -(\sin t)\hat{i} + (\cos t)\hat{j} + (2t^2)\hat{k}$

Calculating the Dot Product $F \cdot dr$

Compute the dot product $F(r(t)) \cdot r'(t)$:

$F \cdot dr = \left[ (-\sin t)\hat{i} + (\cos t)\hat{j} + (2t^2)\hat{k} \right] \cdot \left[ (-\sin t \ \hat{i} + \cos t \ \hat{j} + 4t\hat{k}) dt \right]$

$F \cdot dr = (\sin^2 t + \cos^2 t + 8t^3) dt$

Using the identity $\sin^2 t + \cos^2 t = 1$, this simplifies to:

$F \cdot dr = (1 + 8t^3) dt$

Performing the Integration

Integrate the result from $t=0$ to $t=1$:

$\int_C F \cdot dr = \int_0^1 (1 + 8t^3) dt$

Evaluate the definite integral:

$ \int_0^1 (1 + 8t^3) dt = \left[ t + 8 \frac{t^4}{4} \right]_0^1 $

$ = \left[ t + 2t^4 \right]_0^1 $

$ = (1 + 2(1)^4) - (0 + 2(0)^4) $

$ = (1 + 2) - 0 = 3$

The value of the line integral is 3.