Problem 39 - Thomas Calculus

Question

Surface area Find the area of the surface generated by revolving the arc $$x=t^{2 / 3}, \quad y=t^{2} / 2, \quad 0 \leq t \leq 2$$ about the \(x\) -axis.

Step-by-Step Solution

Answer

The surface area is given by \(\frac{2\pi}{3} \int_0^2 t^{5/3} \sqrt{4 + 9t^{8/3}} \, dt\).

Step 1: Set Up the Surface Area Formula

The formula for the surface area generated by revolving a curve \(y = f(x)\) around the \(x\)-axis is given by: \[S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]In this problem, \(x = t^{2/3}\) and \(y = \frac{t^2}{2}\) parameterize the curve. We need to express everything in terms of \(t\) and find the derivatives.

Step 2: Compute Derivatives

Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):* \(\frac{dx}{dt} = \frac{2}{3}t^{-1/3}\)* \(\frac{dy}{dt} = t\)Next, find \(\left(\frac{dy}{dx}\right)^2\):\[\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t}{\frac{2}{3}t^{-1/3}} = \frac{3t^{4/3}}{2}\]Thus, \[\left(\frac{dy}{dx}\right)^2 = \left(\frac{3t^{4/3}}{2}\right)^2 = \frac{9t^{8/3}}{4}\]

Step 3: Simplify the Expression Inside the Integral

Substitute \(\left(\frac{dy}{dx}\right)^2\) from Step 2 into the integral formula:\[1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{9t^{8/3}}{4}\]Simplifying, we have:\[1 + \frac{9t^{8/3}}{4} = \frac{4 + 9t^{8/3}}{4}\]

Step 4: Substitute into the Surface Area Formula

Substitute \(y = \frac{t^2}{2}\), \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\frac{4 + 9t^{8/3}}{4}}\), \(dx = \frac{2}{3}t^{-1/3} dt\):\[S = \int_0^2 2\pi \left(\frac{t^2}{2}\right) \sqrt{\frac{4 + 9t^{8/3}}{4}} \cdot \frac{2}{3}t^{-1/3} \, dt\]This simplifies to:\[S = \int_0^2 \frac{4\pi t^{5/3}}{3} \sqrt{\frac{4 + 9t^{8/3}}{4}} \, dt\]

Step 5: Simplify and Integrate

Further simplify:\[S = \frac{4\pi}{3} \int_0^2 t^{5/3} \cdot \frac{\sqrt{4 + 9t^{8/3}}}{2} \, dt\]\[S = \frac{2\pi}{3} \int_0^2 t^{5/3} \sqrt{4 + 9t^{8/3}} \, dt\]This integral needs to be evaluated either by substitution or numerical integration methods.

Key Concepts

Parametric Equations

Parametric equations allow us to represent curves in terms of a parameter, often labeled as 't'. This approach can be particularly useful for describing curves that are difficult to express as a single function of two variables, such as complex loops or spirals.
In this particular exercise, the curve is defined by the parametric equations: \

• \( x = t^{2/3} \) describes the horizontal position.
• \( y = \frac{t^2}{2} \) describes the vertical position.

With parametric equations, we can easily calculate derivatives like \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), enabling us to solve problems involving curve length, area, and surface area. It's crucial to understand how to manipulate these kinds of equations because they frequently appear in calculus, particularly in problems involving motion along a curve or geometry of curves.

Revolution Around the X-axis

Revolving a curve around an axis creates a three-dimensional shape or surface. In this exercise, we revolve the described parametric curve around the x-axis.
The surface area of such a shape can be found using specific integral formulas, which account for the circular nature of the revolved section. To calculate the surface area \( S \), we often utilize the integral: \[S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\] This formula stems from the idea of summing up the infinitesimal rings (or disks) that make up the surface when the curve is rotated. The "\(2\pi y\)" term represents the circumference of these discs, while "\( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \)" takes into account the slope of the curve. This whole integral then effectively "sweeps" out the area from start to end points \( a \) to \( b \).

Integration Techniques

Integration is the process of finding the area under a curve or the accumulation of quantities, and it's key to calculating surface areas of revolved solids. In this problem, applying appropriate integration techniques is crucial for correctly finding the surface area.
We start by setting up the integral with the parametric forms and the corresponding transformations. For parametric curves, the expression that we integrate often transforms into a more complex form, involving derivatives and expressions of the parameter like \( t \).
Here, it becomes essential to substitute the parametric derivatives correctly. For our problem at hand, integrating involves:

• Substituting derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) into the formula.
• Simplifying the expression under the square root.
• Handling the integration limits, from \( t = 0 \) to \( t = 2 \).

Sometimes, integrals can lead to complex expressions that require techniques like substitution or numerical methods to solve. Understanding how to manipulate and solve these forms is crucial as they appear frequently in higher-level calculus.

Applications of Derivatives

Derivatives are a fundamental concept in calculus that help us understand how functions change. They are important for determining velocity, acceleration, and, as seen in this problem, allow us to tackle geometric applications such as finding the surface area of a rotated curve.
Taking derivatives of parametric equations involves calculating the rates of change of each parameterized equation, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). These are then used to form the derivative \( \frac{dy}{dx} \), which helps determine the slope of the curve at any point.
For calculating surface areas of revolved solids, derivatives help us to determine:

• How steep the curve is, affecting the circumference of the rings created by rotation.
• The correct transformation of variables through \( \frac{dy}{dx} \).

Accurate computation of these derivatives and integrating them properly contribute to the understanding not only of the geometry but of diverse physical phenomena, demonstrating the wide applicability of calculus.