To find the area enclosed by the curves defined by the parametric equations given, we need to consider the geometry and algebra involved in the problem.

Step 1: Understand the Parametric Equations

The two sets of parametric equations are:

1. $(x_1, y_1, z_1) = (a \cos(t), a \sin(t), ct)$ for $t$ from $0$ to $10\pi$
2. $(x_2, y_2, z_2) = (a \cos(t + g\pi), a \sin(t + g\pi), ct)$ for $t$ from $0$ to $10\pi$

where $a = 1$, $g = \frac{4}{10}$, and $c$ is some constant.

Step 2: Projection onto the XY-plane

Since the area is typically enclosed in a plane, we can project the curves onto the $xy$-plane by ignoring the $z$-coordinate. The projected equations are:

1. $(x_1, y_1) = (\cos(t), \sin(t))$
2. $(x_2, y_2) = (\cos(t + \frac{4\pi}{10}), \sin(t + \frac{4\pi}{10}))$

Step 3: Determine the Enclosed Area

To find the area between these curves in the $xy$-plane over the interval $t$ from $0$ to $10\pi$, we can compute the area between two curves.

Step 4: Use the Parametric Area Formula

The area $A$ between two parametric curves $C_1: (x_1(t), y_1(t))$ and $C_2: (x_2(t), y_2(t))$ from $t = t_1$ to $t = t_2$ is given by:

$A = \frac{1}{2} \int_{t_1}^{t_2} \left[ x_1(t) y_2'(t) - y_1(t) x_2'(t) + x_2(t) y_1'(t) - y_2(t) x_1'(t) \right] \, dt$

Step 5: Calculate the Derivatives

1. For $C_1: (x_1(t), y_1(t)) = (\cos(t), \sin(t))$:

   • $x_1'(t) = -\sin(t)$
   • $y_1'(t) = \cos(t)$

2. For $C_2: (x_2(t), y_2(t)) = (\cos(t + \frac{4\pi}{10}), \sin(t + \frac{4\pi}{10}))$:

   • $x_2'(t) = -\sin(t + \frac{4\pi}{10})$
   • $y_2'(t) = \cos(t + \frac{4\pi}{10})$

Step 6: Substitute into the Formula

Substituting these derivatives into the parametric area formula:

$A = \frac{1}{2} \int_{0}^{10\pi} \left[ \cos(t) \cos(t + \frac{4\pi}{10}) + \sin(t) \sin(t + \frac{4\pi}{10}) - \cos(t + \frac{4\pi}{10}) \cos(t) - \sin(t + \frac{4\pi}{10}) \sin(t) \right] \, dt$

Step 7: Simplify the Integrand

Notice that the terms in the integrand simplify using trigonometric identities, specifically the angle sum identity:

$\cos(t) \cos(t + \frac{4\pi}{10}) + \sin(t) \sin(t + \frac{4\pi}{10}) = \cos\left(t - (t + \frac{4\pi}{10})\right) = \cos\left(-\frac{4\pi}{10}\right)$

$\cos(t + \frac{4\pi}{10}) \cos(t) + \sin(t + \frac{4\pi}{10}) \sin(t) = \cos\left(t + \frac{4\pi}{10} - t\right) = \cos\left(\frac{4\pi}{10}\right)$

The integrand becomes:

$A = \frac{1}{2} \int_{0}^{10\pi} \left[ 2 \cos\left(\frac{4\pi}{10}\right) - 2 \cos\left(\frac{4\pi}{10}\right) \right] \, dt = 0$

Step 8: Interpret the Result

The integral evaluates to zero, indicating that the area enclosed by these curves over the given interval $t$ is zero. This suggests that the curves are symmetric or coincide in such a way that no enclosed area is formed.

Would you like any further details or explanations?

Follow-up Questions:

1. How would the result change if the parameter $g$ were different?
2. Can we consider a different method to find the enclosed area in 3D space?
3. What happens if the curves do not overlap exactly, leading to a non-zero area?
4. Could you analyze the geometric meaning of the parameter $g$ in this context?
5. How does the constant $c$ in the $z$-coordinate affect the problem?

Tip:

Always check for symmetry in parametric curves, as it can often simplify the problem significantly.