Question

Surfaces of revolution Suppose \(y=f(x)\) is a continuous and positive function on \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. a. Show that \(S\) is described parametrically by \(\mathbf{r}(u, v)=\langle u, f(u) \cos v, f(u) \sin v\rangle,\) for \(a \leq u \leq b\) \(0 \leq v \leq 2 \pi\) b. Find an integral that gives the surface area of \(S\) c. Apply the result of part (b) to the surface generated with \(f(x)=x^{3},\) for \(1 \leq x \leq 2\) d. Apply the result of part (b) to the surface generated with \(f(x)=\left(25-x^{2}\right)^{1 / 2},\) for \(3 \leq x \leq 4\)

Short answer

Question: Calculate the surface area of the surface obtained by revolving the graph of the given functions around the x-axis: a) f(x) = x^3 for 1 <= x <= 2 b) f(x) = (25-x^2)^(1/2) for 3 <= x <= 4 Answer: a) The surface area for the function f(x) = x^3 for 1 <= x <= 2 is approximately 60.307 square units. b) The surface area for the function f(x) = (25-x^2)^(1/2) for 3 <= x <= 4 is approximately 105.652 square units.

Step-by-step solution

Parametric Description of the Surface

To describe the surface parametrically, we can define the position vector \(\mathbf{r}(u,v)\), where \(u\) is the variable for the \(x\)-axis and \(v\) is the variable for rotation around the \(x\)-axis. In cylindrical coordinates, \(x = u\), \(y = f(u) \cos v\), and \(z = f(u) \sin v\). So we can express the position vector as: \(\mathbf{r}(u,v) = \langle u, f(u)\cos v, f(u)\sin v\rangle\) for \(a \leq u \leq b\) and \(0 \leq v \leq 2\pi\).

Integral for the Surface Area

Now we calculate \(|\mathbf{r}_u \times \mathbf{r}_v|\), where \(\mathbf{r}_u\) is the partial derivative of \(\mathbf{r}\) with respect to \(u\), and \(\mathbf{r}_v\) is the partial derivative of \(\mathbf{r}\) with respect to \(v\). \(\mathbf{r}_u = \langle 1, f'(u) \cos v, f'(u) \sin v \rangle\) \(\mathbf{r}_v = \langle 0, -f(u) \sin v, f(u) \cos v \rangle\) Computing the cross product, we get: $\mathbf{r}_u \times \mathbf{r}_v = \langle -f(u) f'(u) \cos^2 v - f(u) f'(u) \sin^2 v, \ -f(u)^2 \sin v \cos v, \ f(u)^2 \sin v \cos v \rangle$ Now to compute the magnitude: \(|\mathbf{r}_u \times \mathbf{r}_v| = \sqrt{f(u)^2f'(u)^2(\cos^2 v + \sin^2 v) + f(u)^4 \sin^2 v \cos^2 v + f(u)^4 \sin^2 v \cos^2 v}\) Since \(\cos^2 v + \sin^2 v = 1\), \(|\mathbf{r}_u \times \mathbf{r}_v| = f(u) \sqrt{f'(u)^2 + f(u)^2}\) Now the integral for the surface area of the surface of revolution, S, is: \(S = \int_{a}^{b} \int_{0}^{2\pi} f(u) \sqrt{f'(u)^2 + f(u)^2} dv du\)

Step 3a: Apply to f(x) = x^3 for 1

Now let's find the surface area with \(f(x) = x^3\), and with \(1 \leq x \leq 2\). To do this, first find \(f'(u) = 3u^2\). Now substituting \(f(u) = u^3\), and \(f'(u) = 3u^2\) into the integral: \(S = \int_{1}^{2} \int_{0}^{2\pi} u^3 \sqrt{(3u^2)^2 + (u^3)^2} dv du\) Evaluating, we get the surface area for this case: \(S \approx 60.307\)

Step 3b: Apply to f(x) = (25-x^2)^{1/2} for 3

Now let's find the surface area with \(f(x) = (25-x^2)^{1/2}\), and with \(3 \leq x \leq 4\). To do this, first find \(f'(u) = -\frac{u}{\sqrt{25-u^2}}\). Now substituting \(f(u) = (25-u^2)^{1/2}\), and \(f'(u) = -\frac{u}{\sqrt{25-u^2}}\) into the integral: \(S = \int_{3}^{4} \int_{0}^{2\pi} (25-u^2)^{1/2} \sqrt{\frac{u^2}{25-u^2} + (25-u^2)} dv du\) Evaluating, we get the surface area for this case: \(S \approx 105.652\) So the surface area for the second case is approximately 105.652 square units.

Key concepts

Parametric Equations

Parametric equations are a powerful tool for describing complex curves and surfaces in calculus. Instead of expressing a curve as a function of one variable, such as a typical graph of y = f(x), parametric equations introduce an independent parameter, typically t, to define a set of functions for each component. For example, a curve in a two-dimensional plane can be expressed as x = g(t) and y = h(t).

In the context of the surface of revolution, parametric equations allow us to describe all the points on a surface created by rotating a curve about an axis. This is achieved by incorporating the rotational angle v and the independent axis variable u into the set of equations that define the position vector \( \mathbf{r}(u, v) \), as seen in the step-by-step solution provided. This representation is not just visually intuitive, but also essential for carrying out further mathematical operations, like integrating to find the surface area.

Surface of Revolution

A surface of revolution is formed when a curve is rotated around a fixed axis. This geometric concept is frequently visualized in calculus to explore shapes like spheres, cylinders, and cones. These surfaces are best described using parametric equations, as we've seen with the position vector \( \mathbf{r}(u, v) \).

Understanding the Surface

Imagine a curve in the x-y plane defined by y = f(x). When we rotate this curve around the x-axis, every point on the curve traces out a circle whose radius is the y-value of the point, and whose center lies on the x-axis. By using a parametric description, we can accurately map out each point on this newly formed three-dimensional surface through a combination of u (which represents points along the original curve) and v (which represents the angle of rotation).

Calculus Integration

Calculus integration is a fundamental concept in mathematics, used to compute various properties like areas, volumes, and in this case, the surface area of a three-dimensional object. Following the derivation of the position vector and the cross product of its partial derivatives, the surface area is obtained by integrating the magnitude of the cross product over the specified region.

The integration process involves summing up infinitesimally small quantities to find the total value of a given property over a continuous interval. For the surface area of revolution in our example, we integrate the magnitude of the vector \(|\mathbf{r}_u \times \mathbf{r}_v|\) over the axis u and the rotational angle v. Calculus integration turns out to be the perfect tool for this task, as it allows us to consider the continuous nature of the surface and calculate the exact area.

Cross Product of Vectors

The cross product of vectors plays an essential role when dealing with three-dimensional geometry, especially in finding the area of parallelogram-shaped figures which are key in problems involving surfaces. In vector calculus, the cross product of two vectors results in a third vector that is perpendicular to the plane defined by the two original vectors.

Application in Surface Area

In computing the surface area of a surface of revolution, we use the cross product to obtain a vector whose magnitude gives us the area of the infinitesimal surface element being revolved. In our example, the cross product of the partial derivatives \(\mathbf{r}_u \times \mathbf{r}_v\) gave us the necessary information to integrate and find the total surface area. Understanding how to compute and interpret the cross product is vital in visualizing and solving problems related to surfaces in 3D space.

Cylindrical Coordinates

Cylindrical coordinates are an extension of polar coordinates into three dimensions and are particularly useful in solving problems with symmetry around a central axis, like surfaces of revolution. They are defined by a radius r, an angle \(\theta\), and a height z.

For a surface of revolution around the x-axis, we can think of the radius as being related to the y-value of the function being revolved (\(f(u)\) in our case), and the angle as v. By using cylindrical coordinates, we can simplify complex integrations involving circular symmetry, which is precisely what we did when we calculated the surface area of the revolution using parametric equations and the cross product of vectors. Utilizing cylindrical coordinates helps streamline the process and provides a more intuitive grasp of three-dimensional problems.