To evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$, where:

$\mathbf{F} = (6xy) \mathbf{i} + (6x^2 - 6xy^{'}) \mathbf{j}$

we proceed by parameterizing the three segments of the path $C$:

1. Segment 1: Along the x-axis from $(0,0)$ to $(3,0)$

• Parametric form: $\mathbf{r}_1(t) = (t, 0), \quad 0 \leq t \leq 3$
• $d\mathbf{r}_1 = (1, 0) dt$

Substitute into the dot product:

$\mathbf{F} = (6t \cdot 0)\mathbf{i} + (6t^2 - 6t \cdot 0)\mathbf{j} = (0)\mathbf{i} + (6t^2)\mathbf{j}$

$\mathbf{F} \cdot d\mathbf{r}_1 = 6t^2 \cdot 0 = 0$

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2. Segment 2: Arc of the circle $x^2 + y^2 = 9$ from $(3,0)$ to $(\sqrt{2}, \sqrt{2})$

• Parametric form: $\mathbf{r}_2(\theta) = (3\cos(\theta), 3\sin(\theta))$, with $\theta$ from $0$ to $\frac{\pi}{4}$
• $d\mathbf{r}_2 = (-3\sin(\theta), 3\cos(\theta)) d\theta$

Substitute into the dot product:

$\mathbf{F} = \left(6 \cdot 3\cos(\theta) \cdot 3\sin(\theta)\right)\mathbf{i} + \left(6 \cdot (3\cos(\theta))^2 - 6 \cdot 3\cos(\theta) \cdot 3\sin(\theta)\right)\mathbf{j}$

Simplify the components:

$\mathbf{F} = \left(54\cos(\theta)\sin(\theta)\right)\mathbf{i} + \left(54\cos^2(\theta) - 54\cos(\theta)\sin(\theta)\right)\mathbf{j}$

Dot product:

$\mathbf{F} \cdot d\mathbf{r}_2 = -162\cos(\theta)\sin^2(\theta) + 162\cos^3(\theta)$

Integrate:

$I_2 = \int_0^{\frac{\pi}{4}} (-162 \cos(\theta) \sin^2(\theta) + 162 \cos^3(\theta)) d\theta$

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3. Segment 3: Line from $(\sqrt{2}, \sqrt{2})$ to $(0,0)$

• Parametric form: $\mathbf{r}_3(t) = (\sqrt{2}(1 - t), \sqrt{2}(1 - t))$, with $t \in [0,1]$
• $d\mathbf{r}_3 = (-\sqrt{2}, -\sqrt{2}) dt$

Substitute into the dot product:

$\mathbf{F} = \left(6 \cdot \sqrt{2}(1-t) \cdot \sqrt{2}(1-t)\right)\mathbf{i} + \left(6 (\sqrt{2}(1-t))^2 - 6 \cdot \sqrt{2}(1-t) \cdot \sqrt{2}(1-t)\right)\mathbf{j}$

Simplify:

$\mathbf{F} = 12(1-t)^2 \mathbf{i} + \left(12(1-t)^2 - 12(1-t)^2\right)\mathbf{j} = 12(1-t)^2 \mathbf{i}$

Dot product:

$\mathbf{F} \cdot d\mathbf{r}_3 = -12(1-t)^2 \sqrt{2}$

Integrate:

$I_3 = \int_0^1 -12 \sqrt{2} (1-t)^2 dt$

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Final Result:

$\int_C \mathbf{F} \cdot d\mathbf{r} = I_1 + I_2 + I_3$

Let me calculate these integrals now. sympy import symbols, cos, sin, integrate, pi

Define variables

t, theta = symbols('t theta')

Integral for Segment 2

F_dot_dr2 = -162 * cos(theta) * sin(theta)**2 + 162 * cos(theta)**3 I2 = integrate(F_dot_dr2, (theta, 0, pi/4))

Integral for Segment 3

F_dot_dr3 = -12 * (1 - t)2 * 20.5 I3 = integrate(F_dot_dr3, (t, 0, 1))

I2, I3The evaluated integrals are:
$I_2 = 54\sqrt{2} \quad \text{and} \quad I_3 \approx -5.66$

Since $I_1 = 0$, the total line integral is:

$\int_C \mathbf{F} \cdot d\mathbf{r} \approx 54\sqrt{2} - 5.66 \approx 70.54$

Would you like a more detailed explanation of any specific step?

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Related Questions:

1. How do you parameterize different types of curves for line integrals?
2. What are common methods to evaluate dot products in vector fields?
3. How do you determine limits of integration for parametrized paths?
4. What conditions allow using Green's Theorem instead of direct evaluation?
5. How does the choice of path affect the evaluation of conservative vector fields?

Tip: Always check if a vector field is conservative before performing a line integral—it can simplify calculations using potential functions.