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Chapter 6: Problem 35

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=x, y=2 x, y=6$$

Short Answer

Expert verified
Answer: The volume of the solid generated is \(\frac{243\pi}{4}\).

Step by step solution

01

Find the points of intersection

We need to find the points of intersection of the given curves to determine the limits of integration. Let's find the intersection points of \(y=x\) and \(y=6\). Set \(x = y\) and \(y = 6\). $$x = 6$$ Now let's find the intersection points of \(y=2x\) and \(y=6\). Set \(2x = y\) and \(y = 6\). $$2x = 6$$ $$x = 3$$ So, the region \(R\) is bounded by the points \((3,6)\) and \((6,6)\) along the y-axis.
02

Set up the washer method integral

In the washer method, the volume can be calculated using the formula: $$V = \pi\int_a^b (R(y)^2 - r(y)^2) dy$$ where \(R(y)\) is the outer radius (farther from the \(y\)-axis), and \(r(y)\) is the inner radius (closer to the \(y\)-axis). Since the region is revolved around the y-axis, we will solve both equations for x. We have, $$x = \frac{1}{2}y$$ $$x = y$$ In this case, \(R(y) = y\) (outer radius) and \(r(y) = \frac{1}{2}y\) (inner radius). The limits of integration are from 3 to 6. Plugging these into the formula, we obtain the integral: $$V = \pi\int_3^6 (y^2 - (\frac{1}{2}y)^2) dy$$
03

Simplify the integral

Let's simplify the integrand before integrating. $$V = \pi\int_3^6 (y^2 - \frac{1}{4}y^2) dy$$ $$V = \pi\int_3^6 (\frac{3}{4}y^2) dy$$
04

Integrate and calculate the volume

Now, we'll integrate and calculate the volume of the solid. $$V = \pi\left[\frac{1}{4} (\frac{3}{4}y^3)\right]_3^6$$ $$V = \pi\left[\frac{1}{4} (\frac{3}{4}(6^3) - \frac{3}{4}(3^3))\right]$$ $$V = \pi\left[\frac{1}{4}(324 - 81)\right]$$ $$V = \pi\left[\frac{243}{4}\right]$$ $$V = \frac{243\pi}{4}$$ The volume of the solid generated when the region \(R\) is revolved about the \(y\)-axis is \(\frac{243\pi}{4}\).

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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 straightforward way to find the volume of a solid of revolution. This method is especially useful when the solid is formed by rotating a region around an axis. Imagine you have a flat region that you want to spin around a line. As it spins, it creates a 3-dimensional shape. By visualizing the shape in slices, each slice resembles a circular disk. The volume of each disk can be calculated and summed up to find the total volume of the solid. The formula used is \[ V = \, \pi \, \int_{a}^{b} (r(x)^2) \, dx \]where
  • \(r(x)\) is the radius of each disk,
  • \([a, b]\) are the limits of integration.
The disk method is perfect if the region being rotated fills the entire space between the two axes, without any hollow part in between. It provides a powerful yet simple tool to solve many volume rotation problems.
Washer Method
The washer method extends the disk method by accounting for the hollow parts in the solid. Just like disks, washers are circular, but they have a hole in the center, resembling a donut. This situation arises when there is a gap between the region being revolved and the axis of rotation. The washer method subtracts the volume of this hole from the disk to achieve an accurate volume calculation. The formula used in this method is expressed as: \[ V = \pi \int_{a}^{b} (R(y)^2 - r(y)^2) \, dy \]where
  • \(R(y)\) is the outer radius,
  • \(r(y)\) is the inner radius,
  • \([a, b]\) are the limits of integration.
By subtracting the inner radius squared from the outer radius squared, this method effectively calculates the volume of the washer shaped cross-sections. It's ideal whenever there's an empty space between the axis of rotation and the region, providing an essential technique for more complex situations.
Definite Integrals
Definite integrals are a fundamental tool in calculus, particularly useful in finding areas, volumes, and other quantities that accumulate across an interval. They extend the concept of summing up small changes into a continuous function, allowing for an exact calculation of the accumulated quantity. A definite integral from \(a\) to \(b\) is represented by \[ \int_{a}^{b} f(x) \ dx \]and provides the net area under the curve of \(f(x)\) from \(x = a\) to \(x = b\).
  • \([a, b]\) are the limits of integration, setting the range.
  • \(f(x)\) is the function being integrated, representing the height at each point \(x\).
Definite integrals evaluate this area, factoring in both above and below the x-axis regions. In the context of volume calculations, such as using the disk or washer methods, definite integrals sum up the contributions of each infinitely small slice, providing the complete volume of the solid of revolution.

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Most popular questions from this chapter

A large tank has a plastic window on one wall that is designed to withstand a force of 90,000 N. The square window is \(2 \mathrm{m}\) on a side, and its lower edge is \(1 \mathrm{m}\) from the bottom of the tank. a. If the tank is filled to a depth of \(4 \mathrm{m}\), will the window withstand the resulting force? b. What is the maximum depth to which the tank can be filled without the window failing?

It takes \(100 \mathrm{J}\) of work to stretch a spring \(0.5 \mathrm{m}\) from its equilibrium position. How much work is needed to stretch it an additional \(0.75 \mathrm{m} ?\)

A large building shaped like a box is 50 \(\mathrm{m}\) high with a face that is \(80 \mathrm{m}\) wide. A strong wind blows directly at the face of the building, exerting a pressure of \(150 \mathrm{N} / \mathrm{m}^{2}\) at the ground and increasing with height according to \(P(y)=150+2 y,\) where \(y\) is the height above the ground. Calculate the total force on the building, which is a measure of the resistance that must be included in the design of the building.

a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty .\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}\). b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}\).

A 10-kg mass is attached to a spring that hangs vertically and is stretched 2 m from the equilibrium position of the spring. Assume a linear spring with \(F(x)=k x\) a. How much work is required to compress the spring and lift the mass 0.5 m? b. How much work is required to stretch the spring and lower the mass 0.5 m?
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