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.