Question

Let \(C\) be the curve \(x=f(t)\), \(y=g(t),\) for \(a \leq t \leq b,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous on \([a, b]\) and C does not intersect itself, except possibly at its endpoints. If \(g\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(x\)-axis is $$S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$. Likewise, if \(f\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(y\)-axis is $$S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$ (These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve \(y=f(x)\).) Find the area of the surface obtained by revolving the curve \(x=\cos ^{3} t, y=\sin ^{3} t,\) for \(0 \leq t \leq \pi / 2,\) about the \(x\) -axis.

Short answer

Answer: \(\frac{16\pi}{5}\)

Step-by-step solution

Find the derivatives of f(t) and g(t)

We need to find the derivatives \(f^{\prime}(t)\) and \(g^{\prime}(t)\) for the functions \(f(t) = \cos^3{t}\) and \(g(t) = \sin^3{t}\). We can use the chain rule for this purpose. For \(f(t) = \cos^3{t}\): $$f^{\prime}(t) = 3\cos^2{t}\cdot(-\sin{t})$$ For \(g(t) = \sin^3{t}\): $$g^{\prime}(t) = 3\sin^2{t}\cdot\cos{t}$$

Plug the derivatives into the formula for the surface area

Now we have the derivatives of \(f(t)\) and \(g(t)\). We can plug them into the formula for the surface area of revolution: $$S = \int_{0}^{\frac{\pi}{2}} 2\pi (\sin^3{t}) \sqrt{(-3\sin{t}\cos^2{t})^2 + (3\sin^2{t}\cos{t})^2} dt$$

Simplify the expression inside the square root

Before integrating, let's simplify the expression inside the square root: \((-3\sin{t}\cos^2{t})^2 + (3\sin^2{t}\cos{t})^2 = 9\sin^2{t}\cos^4{t} + 9\sin^4{t}\cos^2{t}\) Now, factor out \(9\sin^2{t}\cos^2{t}\): $$9\sin^2{t}\cos^2{t}(\cos^2{t} + \sin^2{t}) = 9\sin^2{t}\cos^2{t}$$

Integrate the simplified expression

Now we can integrate the simplified expression: $$S = \int_{0}^{\frac{\pi}{2}} 2\pi (\sin^3{t}) \sqrt{9\sin^2{t}\cos^2{t}} dt$$ Factor out constants: $$S = 6\pi \int_{0}^{\frac{\pi}{2}} \sin^3{t} \cdot \sin{t}\cos{t} dt$$ Now, let's change variables: \(u = \cos{t} \Rightarrow du = -\sin{t}dt\) The new bounds for the variable \(u\) are \(u(0) = 1\) and \(u(\frac{\pi}{2}) = 0\). $$S = 6\pi \int_{1}^{0} (1-u^2)^2 du$$ Now, we have a polynomial expression inside the integral. Simply expand the integrand and integrate it term-by-term to find the value of the surface area: $$S = 6\pi \int_{1}^{0} (1 - 2u^2 + u^4) du$$ $$S = 6\pi(-\frac{1}{2} u^2 + \frac{2}{3} u^3 - \frac{1}{5} u^5)\Bigg|^{0}_{1}$$ $$S = 6\pi\left[-\frac{1}{2} + \frac{2}{3} - \frac{1}{5}\right]$$ $$S = 6\pi\cdot\frac{8}{15} = \frac{16\pi}{5}$$

Conclusion

The area of the surface obtained by revolving the curve \(x=\cos^3{t}, y=\sin^3{t}\) for \(0 \leq t \leq \frac{\pi}{2}\) about the \(x\)-axis is \(\frac{16\pi}{5}\).

Key concepts

Calculus

Calculus is a branch of mathematics that focuses on the study of how things change. It provides a framework for modeling systems in which there is change, and a way to deduce the predictions of such models. One of the main tools of calculus is differentiation, which deals with the rate at which quantities change. The other fundamental tool is integration, which is the process of adding up these small changes to determine the total amount of change over a period of time.

In the context of the surface area of revolution, calculus comes into play when we need to find the surface area generated by rotating a curve about an axis. The method involves finding the integral of a certain function which represents the infinitesimal area of the surface element being rotated. This approach allows us to accumulate all these small areas to get the total surface area of the 3-dimensional shape formed by the revolution.

Definite Integrals

Definite integrals are a fundamental concept in calculus which involve integrating a function over a specific interval. The definite integral of a function gives us the net area between the function and the x-axis on a specified interval. This 'net' accounts for areas below the axis being subtractive and areas above being additive. Definite integrals are denoted by the integral sign with limits of integration, indicating the start and end points of the interval over which we integrate.

When calculating the surface area of revolution, we use a definite integral to sum up all the incremental areas as the curve rotates around an axis. This process continues from the starting point 'a' to the ending point 'b' of the curve. The given function inside the integral sign (the integrand) represents the circumference of the infinitesimal disk (when revolving around the x-axis) or the infinitesimal cylindrical shell (when revolving around the y-axis) times the 'thickness' or differential element along the axis of revolution.

Parametric Equations

Parametric equations define a group of quantities as functions of one or more independent variables called parameters. Parametric equations are often used to express the coordinates of the points that make up a geometric object such as a curve or surface. In contrast to Cartesian coordinates, which directly give you the position on a graph using an (x, y) format, parametric equations separate the coordinates into individual components that are defined in terms of a third variable, usually denoted as 't', the parameter.

In a surface area of revolution problem, we use parametric equations to describe the position of a point on a curve that is being revolved. In our exercise, the curve has parametric equations where x and y are defined in terms of 't'. The derivatives of these equations, which are functions of 't', are then used in the formula to calculate the radius and velocity of the rotating curve at any point, essential components for finding the surface area. Using parametric equations and their derivatives in the integral is a powerful way to account for complex curves and movements in calculus.