A curve is given by the following parametric equations. X(T)=2t^2, y(T)=3t, on the interval [0,2]. Find the exact surface area (no decimals) of the solid formed by rotating the parametric curve about the x axis.

To find the surface area of the solid formed by rotating a parametric curve about the \(x\)-axis, we use the following surface area formula for parametric equations:

\[
A = 2\pi \int_{a}^{b} y(t) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
\]

Given parametric equations:
- \(x(t) = 2t^2\)
- \(y(t) = 3t\)
- Interval: \(t \in [0, 2]\)

### Step 1: Compute the derivatives

First, find the derivatives of \(x(t)\) and \(y(t)\):

\[
\frac{dx}{dt} = \frac{d}{dt} (2t^2) = 4t
\]
\[
\frac{dy}{dt} = \frac{d}{dt} (3t) = 3
\]

### Step 2: Substitute into the surface area formula

Now, we substitute these values into the surface area formula. We have:

\[
A = 2\pi \int_{0}^{2} 3t \sqrt{(4t)^2 + 3^2} \, dt
\]

Simplify the expression inside the square root:

\[
A = 2\pi \int_{0}^{2} 3t \sqrt{16t^2 + 9} \, dt
\]

### Step 3: Perform the substitution

Now we solve the integral. Let’s use the substitution \( u = 16t^2 + 9 \), so:

\[
du = 32t \, dt \quad \Rightarrow \quad \frac{du}{32} = t \, dt
\]

When \( t = 0 \), \( u = 9 \), and when \( t = 2 \), \( u = 16(2^2) + 9 = 64 + 9 = 73 \).

Thus, the integral becomes:

\[
A = 2\pi \int_{9}^{73} 3 \cdot \frac{1}{32} \sqrt{u} \, du
\]

Simplify the constants:

\[
A = \frac{3\pi}{16} \int_{9}^{73} \sqrt{u} \, du
\]

### Step 4: Integrate

The integral of \( \sqrt{u} \) is \( \frac{2}{3} u^{3/2} \), so:

\[
A = \frac{3\pi}{16} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{9}^{73}
\]

Simplify the constants:

\[
A = \frac{\pi}{8} \left[ u^{3/2} \right]_{9}^{73}
\]

Now compute the values:

\[
A = \frac{\pi}{8} \left( 73^{3/2} - 9^{3/2} \right)
\]

### Step 5: Calculate \( 73^{3/2} \) and \( 9^{3/2} \)

\[
73^{3/2} = (73^{1/2})^3 = (\sqrt{73})^3
\]
\[
9^{3/2} = (9^{1/2})^3 = (\sqrt{9})^3 = 27
\]

So, the surface area becomes:

\[
A = \frac{\pi}{8} \left( (\sqrt{73})^3 - 27 \right)
\]

This is the exact surface area.

Would you like further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How does the surface area formula differ when rotating around the \(y\)-axis instead of the \(x\)-axis?
2. Can this same process be applied to parametric curves with more complex functions for \(x(t)\) and \(y(t)\)?
3. How would the surface area change if the interval of \(t\) was expanded?
4. How does substitution simplify integration in cases like this?
5. How would you find the surface area if the curve was rotated around a line like \(y = 2\)?

**Tip:** When dealing with parametric curves, always compute the derivatives carefully to avoid errors in the surface area formula.

Find the exact surface area of a solid formed by rotating the parametric curve x(t) = 2t^2, y(t) = 3t on [0,2] around the x-axis. This problem explores parametric equations, calculus, and surface area computation through detailed steps and substitution methods. Suitable for undergraduate calculus students.

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