Problem 59 - Calculus Early Transcendentals

Question

Find the exact area of the surface obtained by rotating the given curve about the \(x\) -axis. $$x=t^{3}, \quad y=t^{2}, \quad 0 \leq t \leq 1$$

Step-by-Step Solution

Answer

The exact surface area is \(\frac{383\pi}{135}\).

Step 1: Set up the Surface Area Formula

To find the surface area of the curve rotated about the \(x\)-axis, we use the formula:\[ S = 2\pi \int_{a}^{b} y \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]In this case, \(a = 0\), \(b = 1\), \(x = t^3\), and \(y = t^2\).

Step 2: Calculate Derivatives

Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).- \(\frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2\)- \(\frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t\)

Step 3: Substitute into Surface Area Formula

Plug the derivatives and \(y\) into the surface area formula:\[ S = 2\pi \int_{0}^{1} t^2 \sqrt{(3t^2)^2 + (2t)^2} \, dt \]This simplifies to:\[ S = 2\pi \int_{0}^{1} t^2 \sqrt{9t^4 + 4t^2} \, dt \]

Step 4: Simplify the Integral

Inside the integral, factor out \(t^2\) from the square root:\[ \sqrt{9t^4 + 4t^2} = \sqrt{t^2(9t^2 + 4)} = t\sqrt{9t^2 + 4} \]So, the integral becomes:\[ S = 2\pi \int_{0}^{1} t^3 \sqrt{9t^2 + 4} \, dt \]

Step 5: Perform the Integration

To integrate \(\int t^3 \sqrt{9t^2 + 4} \, dt\), use substitution:Let \(u = 9t^2 + 4\), then \(\frac{du}{dt} = 18t\) or \(dt = \frac{du}{18t}\). Change the limits accordingly:When \(t = 0\), \(u = 4\), and when \(t = 1\), \(u = 13\).The integral becomes:\[ \int_{4}^{13} \frac{t^3}{18t} \sqrt{u} \, du = \frac{1}{18} \int_{4}^{13} t^2 \sqrt{u} \, du \]Resubstitute \(t^2 = \frac{u - 4}{9}\):\[ S = 2\pi \cdot \frac{1}{18} \int_{4}^{13} \frac{u - 4}{9} \sqrt{u} \, du \]

Step 6: Finish the Integration and Calculate

Solve the integral:\[ \frac{1}{18 * 9} \, \int_{4}^{13} (u^{3/2} - 4u^{1/2}) \, du \]This becomes two integrals:1. \(\int_{4}^{13} u^{3/2} \, du\)2. \(\int_{4}^{13} 4u^{1/2} \, du\)Solve each:- \[ \frac{2}{5}(u^{5/2}) \Big|_{4}^{13} - 4\left(\frac{2}{3}(u^{3/2}) \Big|_{4}^{13}\right) \]Calculate the terms and substitute back into \(S\).

Step 7: Calculate the Result

Evaluate the definite integrals and substitute back for the final area \(S\):- Substitute the limits into the integrated functions to find the exact values.Substitute these back:- \[ 2\pi \frac{1}{18 * 9} \left(\frac{2}{5}((13)^{5/2} - (4)^{5/2}) - \frac{8}{3}((13)^{3/2} - (4)^{3/2})\right) \]Simplify the expression to find the surface area.

Key Concepts

Definite Integral

A definite integral is a powerful mathematical concept used to calculate the area under a curve between two defined points on the horizontal axis. This is particularly useful when the graph of the function is complicated or irregular. In this context, we use definite integrals to find the surface area of a shape that's been rotated around an axis.

When it comes to surface area of revolution problems, our main interest is the "area created" by a curve when it's revolved around an axis, such as the x-axis in this exercise. Using definite integrals, we can encapsulate this concept into a formula, enabling us to calculate the area accurately.

In the exercise, we're interested in the surface area of the curve described by the parametric equations as it's revolved. The general formula for the surface area of revolution is:

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

This integral combines the curve's length element and radius of revolution to encapsulate the surface area into a single expression, allowing for precise calculations.

Parametric Equations

Parametric equations are a way to define a curve by expressing the coordinates as functions of a parameter, commonly denoted by \(t\). This approach is useful when the curve can't easily be described using the standard \(y = f(x)\) method.

In this problem, the curve is defined through parametric equations:

• \(x = t^3\)
• \(y = t^2\)

Here, the variable \(t\) is the parameter, and as \(t\) varies from 0 to 1, it traces out a curve on the plane. Parametric equations can simplify complex curves or paths, like those that loop, overlap, or have segments where traditional graphing methods fail.

For instance, using parametric equations is beneficial in calculating the surface area, as the parameter allows us to easily derive the needed derivatives to substitute into our integral. This technique gives more flexibility and precision in defining and manipulating the curves.

Differentiation of Parametric Form

Differentiating parametric equations is crucial for solving problems involving curves. Since coordinates are defined as functions of a parameter, this approach allows us to find how these coordinates change over that parameter.

For this exercise, we differentiate each parametric equation with respect to \(t\):

• \(\frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2\)
• \(\frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t\)

These derivatives give us the rate of change of \(x\) and \(y\) as \(t\) varies. Incorporating these into the formula for the surface area of revolution allows us to calculate the arc length of the curve as part of determining the surface area.

Through differentiation, we can accurately describe how the curve traces over the interval, enhancing our ability to calculate aspects like surface area and curve length. Differentiation is essential here, not only in calculating rates of change but also in transforming the parametric descriptions into forms usable in integral calculus.