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To find the area of the region bounded by the parabolas \( r = \langle t, 4t^2 \rangle \) and \( r = \langle t, 80 - t^2 \rangle \) for \( -4 \le t \le 4 \) using a line integral, we can use Green's Theorem.

Green's Theorem states:
\[ \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \oint_C \left( P \, dx + Q \, dy \right) \]

For simplicity, let's choose \( P = 0 \) and \( Q = x \), then:
\[ \frac{\partial Q}{\partial x} = 1 \]
\[ \frac{\partial P}{\partial y} = 0 \]

So, the area of the region is:
\[ \text{Area} = \iint_R 1 \, dA = \oint_C x \, dy \]

We need to parameterize the boundary curves. The boundary consists of two curves:

1. \( r_1(t) = \langle t, 4t^2 \rangle \) for \( -4 \le t \le 4 \)
2. \( r_2(t) = \langle t, 80 - t^2 \rangle \) for \( 4 \ge t \ge -4 \)

Now we compute the line integrals over these two segments.

For \( r_1(t) = \langle t, 4t^2 \rangle \):
\[ x = t \]
\[ y = 4t^2 \]
\[ dy = d(4t^2) = 8t \, dt \]
\[ \oint_{C_1} x \, dy = \int_{-4}^{4} t \cdot 8t \, dt = \int_{-4}^{4} 8t^2 \, dt \]

For \( r_2(t) = \langle t, 80 - t^2 \rangle \):
\[ x = t \]
\[ y = 80 - t^2 \]
\[ dy = d(80 - t^2) = -2t \, dt \]
\[ \oint_{C_2} x \, dy = \int_{4}^{-4} t \cdot (-2t) \, dt = -\int_{-4}^{4} 2t^2 \, dt \]

Now we evaluate these integrals:

\[ \oint_{C_1} x \, dy = 8 \int_{-4}^{4} t^2 \, dt = 8 \left[ \frac{t^3}{3} \right]_{-4}^{4} = 8 \left( \frac{64}{3} - \left(-\frac{64}{3}\right) \right) = 8 \cdot \frac{128}{3} = \frac{1024}{3} \]

\[ \oint_{C_2} x \, dy = -2 \int_{-4}^{4} t^2 \, dt = -2 \left[ \frac{t^3}{3} \right]_{-4}^{4} = -2 \cdot \frac{128}{3} = -\frac{256}{3} \]

Adding these integrals gives the total area:
\[ \text{Area} = \frac{1024}{3} + \left(-\frac{256}{3}\right) = \frac{1024}{3} - \frac{256}{3} = \frac{768}{3} = 256 \]

So, the area of the region is \( 256 \).

If you need further details or have any questions, feel free to ask!

Here are 5 questions you might find useful:
1. How does Green's Theorem simplify the computation of areas?
2. What are the parameterizations of the boundary curves used in this problem?
3. Why are \( P = 0 \) and \( Q = x \) chosen for Green's Theorem in this context?
4. How is the trigonometric identity used in integration problems?
5. Can you solve similar problems for different types of regions using Green's Theorem?

**Tip:** When solving integrals over parameterized curves, always ensure that the parameter limits and the function evaluations are carefully checked for consistency.

Learn how to calculate the area of the region bounded by two parabolas using Green's Theorem. Detailed step-by-step solution with parameterization of boundary curves and integration over parameterized segments. Suitable for college-level students and enthusiasts of vector calculus.

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