Problem 20

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Cone frustum The portion of the cone $z=\sqrt{x^{2}+y^{2}} / 3$ between the planes $z=1$ and $z=4 / 3$

Step-by-Step Solution

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Answer

The area is $\frac{7\pi\sqrt{10}}{3}$.

1Step 1: Parameterize the Surface

To express the surface area as a double integral, we first parameterize the surface of the cone frustum. The equation of the cone is given as $z = \frac{\sqrt{x^2 + y^2}}{3}$. By setting $x = r \cos\theta$ and $y = r \sin\theta$, we can express $z = \frac{r}{3}$. Therefore, the parameterization is $\mathbf{r}(r, \theta) = (r \cos\theta, r \sin\theta, \frac{r}{3})$. The parameters satisfy $1 \leq \frac{r}{3} \leq \frac{4}{3}$, which simplifies to $3 \leq r \leq 4$. Also, $\theta$ ranges from $0$ to $2\pi$.

2Step 2: Compute the Partial Derivatives

Next, we compute the partial derivatives of the parameterization $\mathbf{r}(r, \theta)$. - The partial derivative with respect to $r$ is $\mathbf{r}_r = (\cos\theta, \sin\theta, \frac{1}{3})$.- The partial derivative with respect to $\theta$ is $\mathbf{r}_\theta = (-r \sin\theta, r \cos\theta, 0)$.

3Step 3: Compute the Cross Product

To find the normal vector to the surface, compute the cross product $\mathbf{r}_r \times \mathbf{r}_\theta$:\[ \mathbf{r}_r \times \mathbf{r}_\theta = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\cos\theta & \sin\theta & \frac{1}{3} \-r \sin\theta & r \cos\theta & 0\end{array} \right| \]Calculating the determinant, we get:$ \mathbf{r}_r \times \mathbf{r}_\theta = -\frac{r}{3} \cos\theta \mathbf{i} - \frac{r}{3} \sin\theta \mathbf{j} + r \mathbf{k} $.

4Step 4: Compute Magnitude of Cross Product

The magnitude of the cross product gives the magnitude of the normal vector. It is:\[\left\| \mathbf{r}_r \times \mathbf{r}_\theta \right\| = \sqrt{\left(-\frac{r}{3} \cos\theta\right)^2 + \left(-\frac{r}{3} \sin\theta\right)^2 + r^2} = r \sqrt{\frac{1}{9} + 1} = r \sqrt{\frac{10}{9}} = \frac{r \sqrt{10}}{3}\]

5Step 5: Set Up the Double Integral

The surface area $A$ is the double integral over the region defined by $r$ from 3 to 4 and $\theta$ from $0$ to $2\pi$. \[A = \int_0^{2\pi} \int_3^4 \frac{r \sqrt{10}}{3} \, dr \, d\theta\]

6Step 6: Evaluate the Double Integral

We first integrate with respect to $r$:\[\int_3^4 \frac{r \sqrt{10}}{3} \, dr = \frac{\sqrt{10}}{3} \int_3^4 r \, dr = \frac{\sqrt{10}}{3} \left[\frac{r^2}{2}\right]_3^4 = \frac{\sqrt{10}}{3} \left(8 - \frac{9}{2}\right) = \frac{\sqrt{10} \cdot 7}{6}\]Now integrate with respect to $\theta$:\[A = \int_0^{2\pi} \frac{7 \sqrt{10}}{6} \, d\theta = \frac{7 \sqrt{10}}{6} \cdot 2\pi = \frac{7 \sqrt{10} \cdot 2\pi}{6} = \frac{7 \pi \sqrt{10}}{3}\]

7Step 7: Final Answer

Thus, the area of the surface between the planes is $\frac{7 \pi \sqrt{10}}{3}$.

Key Concepts

ParametrizationSurface AreaCone FrustumCross Product

Parametrization

Parametrization is a crucial concept in calculating the surface area of complex shapes, like a cone frustum, using double integrals. By expressing the coordinates in terms of parameters, it simplifies the integration process.
The surface of the cone given by the equation $z = \frac{\sqrt{x^2 + y^2}}{3}$ can be parameterized in terms of $r$ and $\theta$.
We define $x = r \cos\theta$, $y = r \sin\theta$, and $z = \frac{r}{3}$. This allows us to represent the surface using the vector function:

$\mathbf{r}(r, \theta) = (r \cos\theta, r \sin\theta, \frac{r}{3})$

The parameters $r$ and $\theta$ must satisfy the conditions that define our region of integration: $3 \leq r \leq 4$, owing to the boundaries set by the planes $z = 1$ and $z = \frac{4}{3}$, and $0 \leq \theta \leq 2\pi$.
This parametrization effectively converts a complex 3D surface into a form that can be tackled with calculus.

Surface Area

Calculating the surface area of shapes like a cone frustum involves converting integrals over a surface into simpler, double integrals over a region on a plane. The role of parametrization is to make this simplification possible.
The surface area $A$ of the frustum is given by the double integral of the magnitude of the normal vector, found from the cross product of the partial derivatives of the parametric vector function. The setup involves:

Defining the region of integration over the parameters $r$ and $\theta$.
Computing the magnitude of the normal vector to the surface.

Once these elements are defined, the surface area can be expressed as:

$A = \int_0^{2\pi} \int_3^4 \frac{r \sqrt{10}}{3} \, dr \, d\theta$

This integral covers the defined parameter limits, evaluating to give the complete surface area of the specified part of the cone.

Cone Frustum

A cone frustum is a three-dimensional geometric shape derived from a cone with its top cut off by a plane parallel to its base. It retains the circular base and gains a circular top.
In this exercise, the cone is defined by $z = \frac{\sqrt{x^2 + y^2}}{3}$, and the frustum in question is bounded between two planes, $z = 1$ and $z = \frac{4}{3}$.
This transformation creates a frustum with a specific height and varying radii at each end. The process involves measuring the distance along the $z$-axis to find the required parameters, thus simplifying complex surface calculations. By parameterizing the surface, one can effectively compute the area of a portion of this frustum using calculus tools for double integrals.

Cross Product

The cross product is an essential vector operation used to find a vector perpendicular to two other vectors. It's especially useful in calculus for finding normal vectors to surfaces, crucial for surface area integrals.
In the solution, partial derivatives $\mathbf{r}_r$ and $\mathbf{r}_\theta$ are computed from the parameterized function $\mathbf{r}(r, \theta)$: