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Chapter 6: Problem 16

\(A\) 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume that \(x\) and \(y\) are measured in meters. The spherical zone generated when the upper portion of the circle \(x^{2}+y^{2}=100\) on the interval [-8,8] is revolved about the \(x\) -axis

Short Answer

Expert verified
Answer: Approximately 628.87 liters.

Step by step solution

01

Find the radius of the circle

We are given the equation of the circle as \(x^2 + y^2 = 100\). This tells us that the circle's radius is \(\sqrt{100}\), which equals 10 meters.
02

Determine the equation for the volume of the sphere

The volume equation of a sphere is given by: \(V = \frac{4}{3}\pi r^3\)
03

Calculate the volume difference between the two spheres

First, we need to convert 1.5 mm layer of paint to meters. To do this, we utilize the following conversion: $$1.5\,\text{mm} = 0.0015\,\text{m}$$ Next, we calculate the radius of the larger sphere (sphere with paint layer) as: $$r_{1} = 10\,\text{m} + 0.0015\,\text{m} = 10.0015\,\text{m}$$ Now, we calculate the volumes of original sphere (\(V_{1}\)) and larger sphere (\(V_{2}\)). \(V_{1} = \frac{4}{3}\pi (10)^3 = \frac{4000}{3} \pi \,\text{m}^3\) \(V_{2} = \frac{4}{3}\pi (10.0015)^3 = \frac{4006023.3445}{3} \pi\,\text{m}^3\) Calculate the volume difference between the two spheres: $$\Delta V = V_{2} - V_{1} = (\frac{4006023.3445}{3}\pi - \frac{4000}{3}\pi)\,\text{m}^3$$ $$\Delta V \approx 0.2001115\pi\,\text{m}^3$$
04

Find the volume of the paint needed and convert it to liters

First, we need to convert the volume in cubic meters to liters. 1 m鲁 = 1000 L So, the volume of the paint needed in liters would be: $$V_{\text{paint}} = 0.2001115\pi\,\text{m}^3 * 1000\,\text{L/m}^3 = 200.1115\pi\,\text{L}$$ $$V_{\text{paint}} \approx 628.87\,\text{L}$$ Thus, approximately 628.87 liters of paint would be needed to cover the spherical zone.

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Key Concepts

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

Spherical Zone
A spherical zone is a portion of a sphere that lies between two parallel planes which cut through the sphere. In this exercise, we're dealing with the spherical zone created by revolving the segment of a circle. This circular segment is taken from the circle computation:
  • The equation is given by: \( x^2 + y^2 = 100 \).
  • It confines the interval from \( x = -8 \) to \( x = 8 \).
When this section of the circle is rotated around the \( x \)-axis, it creates a three-dimensional shape known as a spherical zone. This concept is crucial when considering partial volumes of spheres or adding attributes like paint layers to a part of the sphere. Understanding the geometry of this zone is essential when calculating added volumes due to coatings like paint.
Volume Calculation
The volume calculation for a spherical zone can be done by finding the difference in the volumes of two different spheres, usually aligned by a small change 鈥� like the addition of a paint layer. Here's how it generally works:
  • Initially, we have a sphere with a known radius. In this case, it's \( 10 \) meters.
  • Add the thickness of the layer 鈥� here, it's a 1.5 mm paint coat, which converts to 0.0015 meters.
  • Calculate the new larger radius by adding the thickness to the original radius, giving us \( 10.0015 \) meters.
With these radii, we proceed to calculate the volume of each sphere using the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). This gives us the volumes before and after the paint layer is applied, and the difference between these volumes gives us the volume of the paint itself.
Layer of Paint
The layer of paint adds an extra dimension to the otherwise plain geometry of a sphere or its zone. Understanding how this layer contributes to the volume teaches us about physical applications of calculus. For this particular exercise:
  • The layer is 1.5 mm thick, which we convert to 0.0015 meters for consistency with radius units.
  • Calculate the total volume of the sphere with the paint added, finding the new volume based on the increased radius.
  • The difference in volume between the original and the coated sphere gives the precise amount of paint needed.
This calculation essentially considers the paint as an ultra-thin hollow shell, adding its own volume to the spherical zone. The exercise also teaches us about practical measurement conversions and their significance in real-world applications like painting large structures.
Revolution About Axis
Revolution about an axis is a fundamental technique in calculus for creating a 3D body from a 2D shape, called a solid of revolution. In this problem, the section of the circle described by \( x^2 + y^2 = 100 \) and constrained within \( x = -8 \) and \( x = 8 \) revolves around the \( x \)-axis to form the spherical zone. When a line segment or shape rotates about an axis:
  • It sweeps out a surface in three-dimensional space.
  • This technique is used to analyze and calculate the properties of complex geometries, including their areas and volumes.
Understanding this method is crucial for disciplines like engineering, where materials and objects often involve these sorts of complex shapes and transformations. By rotating the circle about the \( x \)-axis in this problem, we create a spherical zone impacting the paint volume calculation.

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Most popular questions from this chapter

A rigid body with a mass of \(2 \mathrm{kg}\) moves along a line due to a force that produces a position function \(x(t)=4 t^{2},\) where \(x\) is measured in meters and \(t\) is measured in seconds. Find the work done during the first 5 s in two ways. a. Note that \(x^{\prime \prime}(t)=8 ;\) then use Newton's second law \(\left(F=m a=m x^{\prime \prime}(t)\right)\) to evaluate the work integral \(W=\int_{x_{0}}^{x_{f}} F(x) d x,\) where \(x_{0}\) and \(x_{f}\) are the initial and final positions, respectively. b. Change variables in the work integral and integrate with respect to \(t .\) Be sure your answer agrees with part (a).

Evaluate the following integrals. $$\int 7^{2 x} d x$$

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}(\sinh \ln 3)=\frac{\cosh \ln 3}{3}\) b. \(\frac{d}{d x}(\sinh x)=\cosh x\) and \(\frac{d}{d x}(\cosh x)=-\sinh x\) c. Differentiating the velocity equation for an ocean wave \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)}\) results in the acceleration of the wave. d. \(\ln (1+\sqrt{2})=-\ln (-1+\sqrt{2})\) e. \(\int_{0}^{1} \frac{d x}{4-x^{2}}=\frac{1}{2}\left(\operatorname{coth}^{-1} \frac{1}{2}-\operatorname{coth}^{-1} 0\right)\)

Use l'H么pital's Rule to evaluate the following limits. \(\lim _{x \rightarrow 0^{+}}(\tanh x)^{x}\)

A glass has circular cross sections that taper (linearly) from a radius of \(5 \mathrm{cm}\) at the top of the glass to a radius of \(4 \mathrm{cm}\) at the bottom. The glass is \(15 \mathrm{cm}\) high and full of orange juice. How much work is required to drink all the juice through a straw if your mouth is \(5 \mathrm{cm}\) above the top of the glass? Assume the density of orange juice equals the density of water.
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