Problem 124 - Calculus Volume 3

Question

Find the surface area generated by revolving \(x=t^{2}, y=2 t^{2}, 0 \leq t \leq 1\) about the \(y\) -axis.

Step-by-Step Solution

Answer

The surface area is \(\pi \sqrt{5}\).

Step 1: Parametrize the Curve

We start with the given parametric equations: \(x = t^2\) and \(y = 2t^2\), where \(0 \leq t \leq 1\). These equations represent a curve in the \(xy\)-plane.

Step 2: Find the Surface Area Formula

The formula for the surface area of a surface generated by revolving a curve \(x = f(t)\), \(y = g(t)\) about the \(y\)-axis is given by:\[ A = 2\pi \int_{a}^{b} x \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Step 3: Differentiate the Parametric Equations

Calculate the derivatives: \(\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t\) and \(\frac{dy}{dt} = \frac{d}{dt}(2t^2) = 4t\).

Step 4: Compute the Integral Expression

Substitute \(x = t^2\), \(\frac{dx}{dt} = 2t\), and \(\frac{dy}{dt} = 4t\) into the surface area formula:\[ A = 2\pi \int_{0}^{1} t^2 \sqrt{(2t)^2 + (4t)^2} \, dt \]

Step 5: Simplify the Integrand

Simplify the square root term:\[ \sqrt{(2t)^2 + (4t)^2} = \sqrt{4t^2 + 16t^2} = \sqrt{20t^2} = \sqrt{20} \cdot |t| \] Since \(t \geq 0\) for \(0 \leq t \leq 1\), \(|t| = t\). Thus:\[ A = 2\pi \int_{0}^{1} t^2 \cdot t \cdot \sqrt{20} \, dt = 2\pi \sqrt{20} \int_{0}^{1} t^3 \, dt \]

Step 6: Evaluate the Integral

Evaluate the definite integral:\[ A = 2\pi \sqrt{20} \cdot \left[ \frac{t^4}{4} \right]_{0}^{1} = 2\pi \sqrt{20} \cdot \left( \frac{1^4}{4} - \frac{0^4}{4} \right) \] \[ = 2\pi \sqrt{20} \cdot \frac{1}{4} = \frac{2\pi \sqrt{20}}{4} = \frac{\pi \sqrt{20}}{2} \]

Step 7: Simplify the Final Answer

Simplify \(\sqrt{20}\) to \(2\sqrt{5}\):\[ A = \frac{\pi \times 2\sqrt{5}}{2} = \pi \sqrt{5} \]

Key Concepts

Parametric Equations

Parametric equations are a way to define curves and surfaces with one or more parameters, often represented by the variable \(t\). In this specific problem, the parametric equations \(x = t^2\) and \(y = 2t^2\) describe a curve in the \(xy\)-plane. This means that for each value of \(t\), there is a corresponding point \((x, y)\) on the curve.

• The parameter \(t\) helps us track the motion along the curve as it changes, making it very intuitive to see how the curve is formed.
• Instead of expressing \(y\) in terms of \(x\) explicitly, parametric equations break down the path into steps by relating both \(x\) and \(y\) to \(t\).

This becomes especially useful for more complex curves that are hard to express with a single function \(y = f(x)\), enabling detailed modeling of curves in physics, engineering, and computer graphics.

Integration

Integration is a fundamental concept in calculus that is used to find areas under curves, among many other applications. In this surface area problem, integration helps us to sum up an infinite number of infinitesimally thin disks or rings formed by revolving a curve around the \(y\)-axis.

• The integration process here involves calculating the sum of these ring surfaces by using the integral formula.
• The base of each ring is a small distance along the \(x\)-axis, and its height is determined by the part of the curve being revolved.

Thus, integration in the context of parametric equations involves changes and dimensions captured not through a single function but through the combined change rates of \(x\) and \(y\) with respect to \(t\).

Definite Integral

A definite integral represents the accumulated value of a quantity along an interval \(a, b\). It is used to find the exact area under a curve bounded by certain limits, as demonstrated in the given problem involving the limits \([0, 1]\).

• In our exercise, we use a definite integral from \(t=0\) to \(t=1\) to find the surface area generated by the curve.
• The integral \(\int_{0}^{1} t^3 \, dt\) is calculated to find the total surface by evaluating it over the given interval.

The result of this integral, combined with constants from the surface area formula, presents the surface area of the revolution in precise numerical terms, finally simplified to \(\pi \sqrt{5}\). Such integrals provide an exact value rather than a general idea of growth or area, key to applications requiring precision.

Revolution about the y-axis

Revolution about the \(y\)-axis involves rotating a curve around this axis to generate a three-dimensional surface. In the problem at hand, revolving the curve defined by the parametric equations \(x = t^2\) and \(y = 2t^2\) creates a surface whose area we seek to find.

• This concept is crucial for deriving shapes such as bowls, lamps, and many natural objects that exhibit radial symmetry around the \(y\)-axis.
• The formula for surface area in the context of \(y\)-axis revolution uses \(x\) values, because when revolving around the \(y\)-axis, the curve's movement around this axis is primarily defined by its distance from the \(y\)-axis (which is given by \(x\)).

Thus, understanding the revolution about the \(y\)-axis helps in visualizing and calculating the physical properties of real-world objects and structures based on their geometric attributes.