Chapter 14: Problem 45
Find the area of the given surface. The portion of the paraboloid $$ \mathbf{r}(u, v)=u \cos v \mathbf{i}+u \sin v \mathbf{j}+u^{2} \mathbf{k} $$ for which \(1 \leq u \leq 2,0 \leq v \leq 2 \pi\).
Short Answer
Expert verified
The surface area is \( \frac{\pi}{6}((17)^{3/2} - (5)^{3/2}) \)."
Step by step solution
01
Understand the Problem
The problem requires us to find the area of the surface described by the parametric representation of the paraboloid \( \mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k} \) within specific bounds of the parameters \( u \) and \( v \). We have the bounds as \( 1 \leq u \leq 2 \) and \( 0 \leq v \leq 2\pi \).
02
Set Up the Surface Area Integral
For a parametric surface \( \mathbf{r}(u,v) \), the surface area \( A \) can be found using the formula \( A = \int \int_D \left\| \mathbf{r}_u \times \mathbf{r}_v \right\| \, du \, dv \), where \( \mathbf{r}_u \) and \( \mathbf{r}_v \) are the partial derivatives of \( \mathbf{r} \) with respect to \( u \) and \( v \) respectively, and \( \times \) denotes the cross product.
03
Compute the Partial Derivatives
Calculate the partial derivatives: - \( \mathbf{r}_u = \frac{\partial}{\partial u}(u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k}) = \cos v \mathbf{i} + \sin v \mathbf{j} + 2u \mathbf{k} \).- \( \mathbf{r}_v = \frac{\partial}{\partial v}(u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k}) = -u \sin v \mathbf{i} + u \cos v \mathbf{j} \).
04
Determine the Cross Product
Evaluate \( \mathbf{r}_u \times \mathbf{r}_v \):\[\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cos v & \sin v & 2u \ -u \sin v & u \cos v & 0 \end{vmatrix} = (-2u^2 \cos v) \mathbf{i} + (-2u^2 \sin v) \mathbf{j} + (u) \mathbf{k}.\]
05
Calculate the Magnitude of the Cross Product
The magnitude of the cross product is needed for the surface area integral:\[\left\| \mathbf{r}_u \times \mathbf{r}_v \right\| = \sqrt{(-2u^2 \cos v)^2 + (-2u^2 \sin v)^2 + (u)^2} = \sqrt{4u^4 + u^2} = u\sqrt{4u^2 + 1}.\]
06
Set Up and Evaluate the Integral
Now set up the integral for the surface area:\[A = \int_{0}^{2\pi} \int_{1}^{2} u\sqrt{4u^2 + 1} \, du \, dv.\]First, evaluate the inner integral with respect to \( u \), and then the outer integral with respect to \( v \).
07
Evaluate the Inner Integral
Solve for \( \int_{1}^{2} u\sqrt{4u^2 + 1} \, du \) using substitution: Let \( w = 4u^2 + 1 \), then \( dw = 8u \, du \), and \( du = \frac{1}{8u} dw \). Substituting back gives:\[A_u = \int (1/8)u\sqrt{w} \frac{1}{u} \, dw = \frac{1}{8}\int \sqrt{w} \, dw = \frac{1}{8}\left(\frac{2}{3}w^{3/2}\right) \bigg|_{5}^{17}.\]This evaluates to: \( \frac{1}{12}((17)^{3/2} - (5)^{3/2}) \).
08
Evaluate the Outer Integral
Evaluate the outer integral:\[A_v = \int_{0}^{2\pi} \frac{1}{12}((17)^{3/2} - (5)^{3/2}) \, dv = \frac{1}{12}((17)^{3/2} - (5)^{3/2}) \cdot (2\pi).\]So, the area is \( \frac{2\pi}{12}((17)^{3/2} - (5)^{3/2}) = \frac{\pi}{6}((17)^{3/2} - (5)^{3/2}) \).
09
Write the Final Answer
After calculating, the surface area of the given portion of the paraboloid is \( \frac{\pi}{6}((17)^{3/2} - (5)^{3/2}) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Surfaces
Parametric surfaces are a way to represent surfaces in three-dimensional space using two parameters, often denoted as \(u\) and \(v\). Instead of describing a surface as a function of \(x\) and \(y\), parametric surfaces describe each point on the surface using these parameters to output a position vector \(\mathbf{r}(u, v)\).
This approach allows for the representation of more complex surfaces more naturally, especially those not easily expressible as a function of two variables.
For example, in the given problem, a paraboloid is represented by \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k}\).
Here, \(u\) and \(v\) vary over a given range to map out the surface in space.
This approach allows for the representation of more complex surfaces more naturally, especially those not easily expressible as a function of two variables.
For example, in the given problem, a paraboloid is represented by \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^2 \mathbf{k}\).
Here, \(u\) and \(v\) vary over a given range to map out the surface in space.
- The parameter \(u\) controls the radial distance from the origin, as it affects both the \(x\) and \(y\) components.
- The parameter \(v\) determines the rotation around the z-axis.
Paraboloid
A paraboloid is a three-dimensional shape that resembles a parabola extended in space. It can be classified mainly into two forms: elliptical and hyperbolic. In this exercise, we focus on the elliptical paraboloid, which can be visualized like a dish or a satellite's reflector.
The equation \(z = x^2 + y^2\) describes a standard upright paraboloid. In parametric form, it is often expressed as \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^n \mathbf{k}\), where \(n\) is often \(2\) for the simplest form. Here:
The equation \(z = x^2 + y^2\) describes a standard upright paraboloid. In parametric form, it is often expressed as \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^n \mathbf{k}\), where \(n\) is often \(2\) for the simplest form. Here:
- The ellipse profile is seen in the \(xy\)-plane, with the axis of symmetry along the \(z\)-axis.
- The value of \(z\) increases as \(u\) increases, resulting in a surface that broadens outwards.
Integration Techniques
Integrating over a parametric surface involves certain techniques that distinguish it from simple single-variable calculus. The surface area of a parametric surface is found via a double integral. The key formula utilized is:
\[ A = \int \int_D \left\| \mathbf{r}_u \times \mathbf{r}_v \right\| \, du \, dv \]
Here, \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are partial derivatives with respect to \(u\) and \(v\), which define the tangent vectors to the surface. The cross product \(\mathbf{r}_u \times \mathbf{r}_v\) gives a vector perpendicular to the surface, whose magnitude gives the area element.
The integration process usually involves:
Understanding these techniques requires practice but becomes intuitive with experience, ultimately reinforcing the value of parametrics in elegant mathematical solutions.
\[ A = \int \int_D \left\| \mathbf{r}_u \times \mathbf{r}_v \right\| \, du \, dv \]
Here, \(\mathbf{r}_u\) and \(\mathbf{r}_v\) are partial derivatives with respect to \(u\) and \(v\), which define the tangent vectors to the surface. The cross product \(\mathbf{r}_u \times \mathbf{r}_v\) gives a vector perpendicular to the surface, whose magnitude gives the area element.
The integration process usually involves:
- Finding partial derivatives to get the cross product.
- Calculating the magnitude of the cross product vector.
- Setting up a double integral over the specified parameter domain.
Understanding these techniques requires practice but becomes intuitive with experience, ultimately reinforcing the value of parametrics in elegant mathematical solutions.