x(t) = 3t^2, y(t)=2t, [0,4]. find the exact surface area formed by rotating the parametric curve about the x axis

To find the surface area formed by rotating the parametric curve about the \(x\)-axis, we use the formula for surface area generated by rotating a parametric curve \((x(t), y(t))\) around the \(x\)-axis:

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

Given the parametric equations:

\[
x(t) = 3t^2, \quad y(t) = 2t, \quad t \in [0, 4]
\]

### Step 1: Compute the derivatives
We need to find the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\):

\[
\frac{dx}{dt} = \frac{d}{dt}(3t^2) = 6t
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(2t) = 2
\]

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

Now, we plug these into the surface area formula:

\[
S = 2\pi \int_{0}^{4} y(t) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
\]

Substitute \(y(t) = 2t\), \(\frac{dx}{dt} = 6t\), and \(\frac{dy}{dt} = 2\):

\[
S = 2\pi \int_{0}^{4} 2t \sqrt{(6t)^2 + (2)^2} \, dt
\]
\[
S = 2\pi \int_{0}^{4} 2t \sqrt{36t^2 + 4} \, dt
\]
\[
S = 4\pi \int_{0}^{4} t \sqrt{36t^2 + 4} \, dt
\]

### Step 3: Solve the integral

We can perform a substitution to simplify the integral. Let:

\[
u = 36t^2 + 4
\]
Then, the differential of \(u\) is:

\[
du = 72t \, dt
\]

Now substitute into the integral:

\[
S = 4\pi \int_{u(0)}^{u(4)} \frac{\sqrt{u}}{72} \, du
\]
\[
S = \frac{4\pi}{72} \int_{4}^{580} \sqrt{u} \, du
\]
\[
S = \frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_{4}^{580}
\]
\[
S = \frac{\pi}{18} \left( \frac{2}{3} \left(580^{3/2} - 4^{3/2}\right) \right)
\]
\[
S = \frac{\pi}{18} \left( \frac{2}{3} \left(580\sqrt{580} - 8\right) \right)
\]

### Step 4: Final calculation
Now, calculating the exact value of the expression:

\[
S = \frac{2\pi}{54} \left(580\sqrt{580} - 8\right)
\]

This represents the exact surface area formed by rotating the curve about the \(x\)-axis.

---

Would you like more details on the simplifications or approximations?

Here are some related questions:

1. How do you find the surface area of a curve rotated about the \(y\)-axis?
2. What is the difference between parametric and polar coordinates in surface area calculations?
3. Can you find the volume generated by rotating the same curve about the \(x\)-axis?
4. How does the surface area formula change if the curve is rotated about the \(y\)-axis?
5. How would you compute the surface area if the curve was rotated about another arbitrary axis?

**Tip:** In parametric equations, always remember to differentiate correctly with respect to the parameter \(t\), as errors in differentiation can affect the final outcome of integral calculations.

x(t) = 3t^2, y(t) = 2t, [0,4]. Find the exact surface area formed by rotating the parametric curve about the x-axis.

Learn how to compute the surface area formed by rotating the parametric curve x(t) = 3t^2, y(t) = 2t, t ∈ [0, 4] about the x-axis. This solution uses parametric equations and integral calculus to find the exact surface area. Ideal for advanced high school or undergraduate calculus students.

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