Question

For the following exercises, find the area of the surface obtained by rotating the given curve about the \(x\) -axis\(x=t^{3}, \quad y=t^{2}, \quad 0 \leq t \leq 1\)

Short answer

The integral gives the area of the surface when solved, typically using numerical integration or specialized functions.

Step-by-step solution

Parametrize the surface

The given curve is parametrized by \(x = t^3\) and \(y = t^2\). This describes a curve in parametrized form where both \(x\) and \(y\) are expressed in terms of a parameter \(t\).

Calculate the derivatives

We'll find the derivatives of \(x\) and \(y\) with respect to \(t\). Compute \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 2t \).

Find the differential arc length

The differential arc length in parametric form is \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Substituting the derivatives, \(ds = \sqrt{(3t^2)^2 + (2t)^2} \, dt = \sqrt{9t^4 + 4t^2} \, dt\).

Expression for the surface area

The surface area \(S\) of a surface of revolution about the x-axis is given by \(S = \int 2\pi y \, ds\). With \(y = t^2\), the surface area expression becomes \(S = \int_{0}^{1} 2\pi(t^2) \sqrt{9t^4 + 4t^2} \, dt\).

Evaluate the integral

To find the surface area, evaluate the definite integral. Simplify \(\sqrt{9t^4 + 4t^2}\) first. Factor \(t^2\) out: \(t \sqrt{9t^2 + 4}\). The integral becomes \(S = \int_{0}^{1} 2\pi t^3 \sqrt{9t^2 + 4} \, dt\). Using substitution or numerical integration, solve this to get the precise value.

Key concepts

Parametric Equations

Parametric equations provide a unique way to describe a curve by expressing the coordinates in terms of a third parameter, usually denoted by \( t \).
This approach is especially useful when you want to represent complex curves that aren't easily described by a single function of \( x \) or \( y \) alone.

The concept can be visualized as:

• Each curve point is associated with a specific parameter value.
• Both \( x \) and \( y \) are functions of \( t \).
• The curve is traced as \( t \) varies over a specified range.

In the original problem, we used the parametric equations \( x = t^3 \) and \( y = t^2 \) to describe the curve.
As \( t \) varies from 0 to 1, these equations generate each point along the curve in the \( xy \)-plane.

Integral Calculus

Integral calculus is essential for calculating areas and other quantities that accumulate as you move along a curve.
In the context of surface areas of revolution, integrals help sum infinitesimal pieces of surface area to find the total area.

Here's what integral calculus contributes:

• Breaking down a complex surface into smaller, manageable segments.
• Summing these segments to get a total measurement, like area or volume.
• In our exercise: Finding the surface area using an integral of the form \( S = \int f(t) \, dt \).

The exercise required evaluating an integral to compute the surface area of the given curve.
Here, definite integrals combined with the derivatives of parametric equations help relate each segment of the curve to an increment of area.

Surface Integrals

Surface integrals extend the concepts of integration from lines to two-dimensional surfaces, specifically focusing on accumulative properties like area over surfaces in a given space.
For surface areas of revolution, these integrals include a rotating action around an axis, collecting surfaces from sections of a curve.

Consider:

• The surface area created by rotating a curve around an axis represents a practical application of surface integrals.
• The process involves considering infinitesimally small strips of the curve and integrating these around the axis of rotation.
• In our problem, we evaluated the surface integral \( S = \int_{0}^{1} 2\pi y \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \).

This integral accounts for every tiny wrapped piece of the surface, precisely measuring the total area formed.

Curve Parametrization

Parametrizing a curve involves expressing both \( x \) and \( y \) as functions of an independent parameter, simplifying calculations for complex shapes.
It enables the translation of a curve into a form that is easier to manipulate algebraically, particularly useful for surfaces of revolution.

Advantages of Parametrization:

• Simplifies computation of derivatives needed for integral calculations.
• Accurately accounts for the piecewise makeup of curves not defined easily by standard functions.
• In this exercise, \( x = t^3 \) and \( y = t^2 \) formed our parametrization for the curve.

Curve parametrization is a foundational step in solving surface area problems involving rotational surfaces, providing the building blocks for derivatives and integrals.