Question

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

Short answer

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

Step-by-step solution

Parametrize the curve

The given parametric equations are \(x = t^2\) and \(y = 2t^2\). It describes a curve in the plane with \(t\) as the parameter.

Differentiate with respect to the parameter

To find the derivatives, calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Here, \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 4t \).

Determine bounds and setup formula for surface area

The surface area \( A \) of a surface generated by revolving a parametric curve \( (x(t), y(t)) \) about the \(y\)-axis is given by the integral: \[ A = 2\pi \int_a^b x(t) \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \]Here, set \( a = 0 \) and \( b = 1 \).

Plug in the parametric equations and derivatives

Plug the values \(x(t) = t^2\), \( \frac{dx}{dt} = 2t \), and \( \frac{dy}{dt} = 4t \) into the surface area formula:\[ A = 2\pi \int_0^1 t^2 \sqrt{(2t)^2 + (4t)^2} \, dt \]Simplify inside the square root: \[ A = 2\pi \int_0^1 t^2 \sqrt{4t^2 + 16t^2} \, dt = 2\pi \int_0^1 t^2 \sqrt{20t^2} \, dt \]Thus,\[ A = 2\pi \int_0^1 t^2 \cdot \sqrt{20} \, t \, dt = 2\pi \sqrt{20} \int_0^1 t^3 \, dt \]

Integrate

Compute the integral:\[ \int_0^1 t^3 \, dt = \left. \frac{t^4}{4} \right|_0^1 = \frac{1}{4} \]Multiply with the constant:\[ A = 2\pi \sqrt{20} \cdot \frac{1}{4} \]

Simplify the expression

Simplify the expression:\[ \sqrt{20} = 2\sqrt{5} \]Combine:\[ A = 2\pi \cdot 2\sqrt{5} \cdot \frac{1}{4} \]\[ A = \frac{\pi \sqrt{5}}{\sqrt{5}} = \pi \sqrt{5} \]

Conclude the solution

The surface area generated by revolving the curve about the y-axis is \( \pi \sqrt{5} \).

Key concepts

Parametric Equations

Parametric equations offer an elegant method to describe curves or surfaces using parameters instead of relying solely on variables like "x" and "y". When dealing with parametric equations, each coordinate (x and y) is expressed as a function of a third parameter, often denoted as "t". This parameter "t" maintains a specific range to define a curve segment.

In our problem, we have the parametric equations:

• \(x = t^2\)
• \(y = 2t^2\)

These equations transform a simple parameter "t" into a position on a two-dimensional plane. The range of the parameter "t" is often given, here it's from 0 to 1. This parameterization is useful in various mathematical and engineering problems as it offers flexibility and improved handling of curves that aren't easily described by a single function.

By discussing curves parametrically, we benefit from literally 'charting' paths that aren't limited by the constraints of traditional functions. This leads to more efficient calculations and analyses, such as determining points, tangents, and, in this case, surfaces generated by revolutions.

Surface Area of Revolution

The surface area of revolution is the area of a three-dimensional shape created as a two-dimensional curve revolves around an axis. To calculate such surfaces, we often utilize calculus methods - commonly constructing integrals based on curves defined by parametric equations.

In this exercise, the parametric curve revolves around the y-axis. To determine the surface area, the integral utilizes both the parametric function and its derivatives. The standard formula for the surface area "A" when the curve \((x(t), y(t))\) revolves around the y-axis is:

\[ A = 2\pi \int_a^b x(t) \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \]This integral sums up all the infinitesimally small strips around the curve, resembling surfaces of circular discs laid end-to-end. It captures the length and breadth of each slice (dependent on the derivatives of x and y concerning "t") and piles them up over the range of the parameter "t".

This extends our understanding of calculus beyond just the surface, to a tangible structure - a real-world application of theoretical principles.

Integration Techniques

Integration, the reverse process of differentiation, plays a crucial role in calculating the surface area of revolution for parametric curves. The integral slices these curves into minute elements, compounding them into either a single value or function depending on parameters.

From the provided solution, integration was utilized to process the expression stemming from the parametric equations and their derivatives:

\[ A = 2\pi \sqrt{20} \int_0^1 t^3 \, dt \]To solve this, first simplify under the square root, then reset the focus to evaluating the integral, \( \int_0^1 t^3 \, dt = \frac{t^4}{4} \Big|_0^1 \).For such integrals, apply the fundamental theorem of calculus, equating the antiderivative evaluated at boundaries "a" and "b" (0 and 1 in this case). Understanding these steps is pivotal, requiring repeated practice. Integration is essential not just for geometry, but across physics and engineering, playing a vital role in obtaining quantities like areas, volumes, and other metrics from dynamic systems.The process highlights how integration condenses infinite behaviors into singular expressions, forming the backbone of continuous methods in science and technology.