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Chapter 10: Problem 75

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

Short Answer

Expert verified
The surface area is \( 14\pi \sqrt{13} - 18\pi \).

Step by step solution

01

Write down the formula for the surface area of revolution

The formula to find the surface area of a curve revolved around the x-axis over an interval \([a, b]\) is given by: \[ A = 2\pi \int_{a}^{b} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] where \( y = 3t \) and \( 0 \leq t \leq 2 \).
02

Compute the derivatives

Calculate the derivatives of \(x\) and \(y\) with respect to \(t\). Start with \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = t^2 \), \( \frac{dx}{dt} = 2t \). For \( y = 3t \), \( \frac{dy}{dt} = 3 \).
03

Setup the integrand

Substitute the derivatives into the square root part of the integral and simplify. We have: \[ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(2t)^2 + 3^2} = \sqrt{4t^2 + 9} \] The integrand becomes: \( 2\pi (3t) \cdot \sqrt{4t^2 + 9} \).
04

Integrate the expression

Calculate the definite integral: \[ A = 2\pi \int_{0}^{2} 3t \sqrt{4t^2 + 9} \, dt \] Use a substitution method, such as \( u = 4t^2 + 9 \), to solve the integral. Calculate the change of limits and solve the integral to find the area.
05

Evaluate the integral

After performing integration by substitution or by using integral tables to solve \( \int 3t \sqrt{4t^2 + 9} \, dt \), evaluate the definite integral from 0 to 2. This yields: \[ A = 2\pi \left[ \frac{1}{6}(4t^2 + 9)^{3/2} \right]_{0}^{2} \] Compute the final result by plugging the limits into the expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is all about understanding rates of change and accumulation of quantities. Surface area of revolution is one such application. By revolving a curve around an axis, usually the x-axis or y-axis, we create a 3D surface. Calculus helps us find out how much 'space' that surface takes up.
To do this, we need to think about the curve as a set of tiny line segments. As we spin them around the axis, they create circular strips. The surface area is then the sum of all these strips. The key tool here is integration, which allows us to add up an infinite number of these small areas to find the total surface area.
So, calculus is crucial because it provides the method, via integration, to calculate the total surface area from a given set of conditions and equations.
Definite Integral
The definite integral is a mathematical concept used to find the total accumulation of a quantity. It's particularly useful in this exercise to compute surface areas. The definite integral has both a start (\(a\)) and stop (\(b\)) point. In our problem, these are 0 and 2, representing the interval of the parameter \(t\).
Integrals help us to sum an infinite number of infintesimally small products, which, in this scenario, are the bits of the curve forming the surface. The formula we use connects a function, in our case, representing the motion around the x-axis, to the integral. A crucial part is the choice of the correct function for integration. Here it's \(3t \sqrt{4t^2 + 9}\) based on our parametric equations.
Ultimately, evaluating the definite integral gives us a specific numerical result鈥攚hich is the calculated total surface area in our problem.
Parametric Curves
Parametric curves use parameters to define equations for curves, allowing us to express complex curves in a simple form. In this exercise, we use a parameter \(t\) to define our curve as \(x = t^2\) and \(y = 3t\). This represents a path in which 'time' \(t\) illustrates the position of the point on the curve.
Using parametric equations offers flexibility. We are not limited to functions that can be expressed as \(y\) in terms of \(x\), or vice versa. This is especially useful in surface area calculations of revolutions, where calculating derivatives of parametric equations is straightforward, aiding in setting up our integral.
In essence, parametric curves provide a valuable way to handle complex shapes and derive necessary components to solve problems in calculus, like our task of determining the surface area of a revolution.

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Most popular questions from this chapter

Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix. $$9 x^{2}-24 x y+16 y^{2}-80 x-60 y+100=0$$

Find an equation for a hyperbola that satisfies the given conditions. [Note: In some cases there may be more than one hyperbola.] (a) Vertices \((\pm 2,0) ;\) foci \((\pm 3,0)\) (b) Vertices \((0, \pm 2) ;\) asymptotes \(y=\pm \frac{2}{3} x\)

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5. ] Use the formulas above in these exercises. Find the area of the surface generated by revolving \(x=6 t\) \(y=4 t^{2}(0 \leq t \leq 1)\) about the \(y\) -axis.

Find a polar equation for the ellipse that has its focus at the pole and satisfies the stated conditions. $$ \begin{array}{l}{\text { (a) Directrix to the right of the pole; } a=8 ; e=\frac{1}{2}} \\ {\text { (b) Directrix below the pole; } a=4 ; e=\frac{3}{5}}\end{array} $$

Vertical and horizontal asymptotes of polar curves can sometimes be detected by investigating the behavior of \(x=r \cos \theta\) and \(y=r \sin \theta\) as \(\theta\) varies. This idea is used in these exercises. Show that the hyperbolic spiral \(r=1 / \theta(\theta>0)\) has a horizontal asymptote at \(y=1\) by showing that \(y \rightarrow 1\) and \(x \rightarrow+\infty\) as \(\theta \rightarrow 0^{+} .\) Confirm this result by generating the spiral with a graphing utility.
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