Find the surface area generated by rotating the given curve about the y-axis.
x = 9t^2,    y = 6t^3,    0 ≤ t ≤ 5

To find the surface area generated by rotating the curve around the \( y \)-axis, we can use the formula for the surface area of a curve rotated about the \( y \)-axis:

\[
A = 2\pi \int_{y_1}^{y_2} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy
\]

However, since the given curve is parameterized by \( t \), with:

\[
x = 9t^2 \quad \text{and} \quad y = 6t^3
\]

we will modify the formula for parametric equations. The surface area for a curve rotated around the \( y \)-axis is given by:

\[
A = 2\pi \int_{t_1}^{t_2} x \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
\]

### Step 1: Compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

\[
\frac{dx}{dt} = \frac{d}{dt} (9t^2) = 18t
\]
\[
\frac{dy}{dt} = \frac{d}{dt} (6t^3) = 18t^2
\]

### Step 2: Plug these into the formula

The surface area is now:

\[
A = 2\pi \int_{0}^{5} (9t^2) \sqrt{(18t)^2 + (18t^2)^2} \, dt
\]

Simplifying the expression inside the square root:

\[
A = 2\pi \int_{0}^{5} (9t^2) \sqrt{324t^2 + 324t^4} \, dt
\]
\[
A = 2\pi \int_{0}^{5} (9t^2) \sqrt{324t^2 (1 + t^2)} \, dt
\]
\[
A = 2\pi \int_{0}^{5} (9t^2) \cdot 18t \sqrt{1 + t^2} \, dt
\]
\[
A = 324\pi \int_{0}^{5} t^3 \sqrt{1 + t^2} \, dt
\]

### Step 3: Solve the integral

Let \( u = 1 + t^2 \), then \( du = 2t \, dt \). Changing the limits accordingly:

When \( t = 0 \), \( u = 1 \).

When \( t = 5 \), \( u = 26 \).

Now rewrite the integral:

\[
A = 324\pi \int_{1}^{26} \frac{(u - 1)}{2} \sqrt{u} \, du
\]

At this point we would apply integration by substitution to solve for the surface area.

Let me complete the calculation for you:The surface area generated by rotating the given curve about the \( y \)-axis is approximately:

\[
A \approx 135.72 + 128,795.14 \cdot \sqrt{26} \quad \text{square units}
\]

This expression combines a constant part and a term involving \( \sqrt{26} \).

Do you want more details on this calculation or have any specific questions?

Here are 5 related questions to expand your understanding:
1. How does the method for finding surface area change when rotating around the \(x\)-axis instead of the \(y\)-axis?
2. What are the steps to apply parameterization to curves in other surface area problems?
3. How would the formula differ for non-parametric curves when rotated about the \(y\)-axis?
4. Can this method be used for any curve, or are there conditions that must be met?
5. What geometric interpretation can you provide for the surface generated in this specific problem?

**Tip:** When integrating parametric curves, ensure that the limits of integration correspond to the parameter's range, not the \(x\)- or \(y\)-values directly.

Find the surface area generated by rotating the given curve about the y-axis. x = 9t^2, y = 6t^3, 0 ≤ t ≤ 5

Learn how to calculate the surface area of a parametric curve rotated about the y-axis. This example involves the parametric equations x = 9t² and y = 6t³ over the interval 0 ≤ t ≤ 5. The solution uses the method of parametric integration and the formula for surface area of revolution. Suitable for Grades 11-12 and college-level calculus students.

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