Math Problem Statement

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

Solution

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

A=2πy1y2x1+(dxdy)2dyA = 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 tt, with:

x=9t2andy=6t3x = 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 yy-axis is given by:

A=2πt1t2x(dxdt)2+(dydt)2dtA = 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 dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}

dxdt=ddt(9t2)=18t\frac{dx}{dt} = \frac{d}{dt} (9t^2) = 18t dydt=ddt(6t3)=18t2\frac{dy}{dt} = \frac{d}{dt} (6t^3) = 18t^2

Step 2: Plug these into the formula

The surface area is now:

A=2π05(9t2)(18t)2+(18t2)2dtA = 2\pi \int_{0}^{5} (9t^2) \sqrt{(18t)^2 + (18t^2)^2} \, dt

Simplifying the expression inside the square root:

A=2π05(9t2)324t2+324t4dtA = 2\pi \int_{0}^{5} (9t^2) \sqrt{324t^2 + 324t^4} \, dt A=2π05(9t2)324t2(1+t2)dtA = 2\pi \int_{0}^{5} (9t^2) \sqrt{324t^2 (1 + t^2)} \, dt A=2π05(9t2)18t1+t2dtA = 2\pi \int_{0}^{5} (9t^2) \cdot 18t \sqrt{1 + t^2} \, dt A=324π05t31+t2dtA = 324\pi \int_{0}^{5} t^3 \sqrt{1 + t^2} \, dt

Step 3: Solve the integral

Let u=1+t2u = 1 + t^2, then du=2tdtdu = 2t \, dt. Changing the limits accordingly:

When t=0t = 0, u=1u = 1.

When t=5t = 5, u=26u = 26.

Now rewrite the integral:

A=324π126(u1)2uduA = 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 yy-axis is approximately:

A135.72+128,795.1426square unitsA \approx 135.72 + 128,795.14 \cdot \sqrt{26} \quad \text{square units}

This expression combines a constant part and a term involving 26\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 xx-axis instead of the yy-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 yy-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 xx- or yy-values directly.

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Math Problem Analysis

Mathematical Concepts

Surface Area of a Solid of Revolution
Parametric Equations
Integral Calculus

Formulas

Surface area of a curve rotated about the y-axis: A = 2π ∫ x √(1 + (dx/dy)²) dy
Parametric surface area formula: A = 2π ∫ x √((dx/dt)² + (dy/dt)²) dt

Theorems

Method of Parametric Integration
Integration by Substitution

Suitable Grade Level

Grades 11-12 and College Calculus