/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Designing a wok You are designin... [FREE SOLUTION] | 91Ó°ÊÓ

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Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about 3 L if you make it \(9 \mathrm{cm}\) deep and give the sphere a radius of 16 \(\mathrm{cm} .\) To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? \(\left(1 \mathrm{L}=1000 \mathrm{cm}^{3}\right)\)

Short Answer

Expert verified
The volume of the wok is approximately 3305 cm³.

Step by step solution

01

Set up the Volume Integral

The wok is visualized as a spherical cap, which is a segment of a sphere. The volume of this cap can be calculated using integration. The equation of the sphere in terms of its radius, 16 cm, is \( x^2 + y^2 = 16^2 \). We revolve the portion of the sphere from \( y = -16 \) to \( y = -7 \) about the y-axis to find the volume (since it is 9 cm deep, starting 7 cm from the center).
02

Solve for Radius of Revolution

To express the volume of revolution, express \( x \) in terms of \( y \). From the sphere's equation, \( x^2 = 256 - y^2 \), giving \( x = \sqrt{256 - y^2} \). This will be the radius of the disk at height \( y \).
03

Set Up the Definite Integral

The volume of the wok is calculated by revolving the curve from \( y = -16 \) to \( y = -7 \) using the disk method. The formula for the volume \( V \) is: \[ V = \pi \int_{-16}^{-7} (\sqrt{256 - y^2})^2 \, dy. \] This simplifies to: \[ V = \pi \int_{-16}^{-7} (256 - y^2) \, dy. \]
04

Calculate the Integral

Evaluate the integral \( V = \pi \int_{-16}^{-7} (256 - y^2) \, dy \): 1. Compute the antiderivative: \( 256y - \frac{y^3}{3} \).2. Evaluate at limits: \[ \Bigg[ 256y - \frac{y^3}{3} \Bigg]_{-16}^{-7}. \]3. Plug in values: - For \( y = -7 \): \( 256(-7) - \frac{(-7)^3}{3} = -1792 + \frac{343}{3} \) - For \( y = -16 \): \( 256(-16) - \frac{(-16)^3}{3} = -4096 + \frac{4096}{3} \)
05

Simplify and Solve the Integral

Complete calculations for subtracting: \[ V = \pi \left[-1792 + \frac{343}{3} + 4096 - \frac{4096}{3} \right]. \]Simplify inside brackets and calculate: \( -1792 + \frac{343}{3} + 4096 - \frac{4096}{3} = 2304 - \frac{3753}{3}.\)Continue simplifying: \( 2304 - 1251 = 1053. \)Thus, \( V = \pi \times 1053. \)
06

Compute Volume Estimate

Taking \( \pi \approx 3.14 \), estimate the volume:\[ V \approx 3.14 \times 1053. = 3305.42 \, \text{cm}^3.\]Round to the nearest cubic centimeter:\( V \approx 3305 \, \text{cm}^3. \)

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Key Concepts

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

Disk Method
The disk method is a popular technique in integral calculus for finding the volume of a solid of revolution. It is particularly useful when you want to find the volume of a solid formed by revolving a 2D shape around an axis. In our wok problem, we revolve a segment of a sphere around the y-axis.

Imagine slicing this solid perpendicular to the axis of rotation into thin disks. These disks have a small thickness \( dy \) and a radius \( x \), which changes as you move along the axis. The formula for the volume of a disk is \( \pi x^2 \), where \( x \) is the radius of the disk.

To find the total volume, integrate the volume of each of these tiny disks over the desired interval. The integral for the wok, revolving from \( y = -16 \) to \( y = -7 \), becomes:
  • \[ V = \pi \int_{-16}^{-7} (\sqrt{256 - y^2})^2 \, dy. \]
  • This simplifies to \( V = \pi \int_{-16}^{-7} (256 - y^2) \, dy. \)
By calculating this integral, you sum up all the infinitesimally small disks to find the total volume of the wok.
Integral Calculus
Integral calculus is essential for computing areas, volumes, and other quantities that require accumulation of values. In the context of our wok design, integral calculus helps us calculate the volume of a complex solid shape that does not have a simple geometric formula.

We rely on integrals to find the exact volume by considering it as the sum of infinitesimal elements. The integral in the wok problem is set up using the disk method:

The equation is:
  • \[ V = \pi \int_{-16}^{-7} (256 - y^2) \, dy. \]
To solve this, we find the antiderivative, which is \( 256y - \frac{y^3}{3} \). Evaluating this expression at the limits \( y = -16 \) and \( y = -7 \) gives us the exact volume. Integral calculus provides a way to compute these quantities which would otherwise require complicated measurements.
Solid of Revolution
The problem involving the wok is a great example of a solid of revolution. When a shape is rotated around an axis to create a 3D object, it becomes a solid of revolution. Our wok is visualized as part of a sphere, which becomes solid when the sphere's cap is revolved around the y-axis.

In many practical situations like designing cooking utensils or machine parts, understanding solids of revolution is crucial since this process allows transforming 2D designs into useful 3D objects.

The shape of the wok is derived by using the concept that the spherical bowl is a cap of a sphere, thus forming a solid of revolution. Understanding the characteristics of the base shape, and how it transforms when revolved, is essential in calculating properties like surface area and volume using methods such as the disk method. This approach ensures that the design will hold the desired volume efficiently.

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Most popular questions from this chapter

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$x=y^{3} / 3, \quad 0 \leq y \leq 1 ; \quad y \text { -axis }$$

The region in the first quadrant that is bounded above by the curve \(y=1 / \sqrt{x}\), on the left by the line \(x=1 / 4,\) and below by the line \(y=1\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid by a. the washer method. b. the shell method.

Find the volume of the solid generated by revolving each region about the \(y\) -axis. The region in the first quadrant bounded on the left by the circle \(x^{2}+y^{2}=3,\) on the right by the line \(x=\sqrt{3},\) and above by the line \(y=\sqrt{3}\)

Use a CAS to perform the following steps for the given graph of the function over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure 6.22.) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$f(x)=x^{1 / 3}+x^{2 / 3}, \quad 0 \leq x \leq 2$$

Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=2\).

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