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Chapter 8: Problem 22

Let \(a, b \in \mathbb{R}\) with \(0 \leq a

Short Answer

Expert verified
In this problem, we are given a planar region R bounded by the lines 饾懃=饾憥, 饾懃=饾憦, and with the length function 饾憴(饾懃). We aim to find the volume of the solid formed by revolving this region around the y-axis. To do this, we approximate the volume of each cylindrical disk formed by revolving a small length d饾懃 in the region R and sum them up. Thus, the volume can be written as an integral over the interval [饾憥,饾憦]: \( V = \int_{a}^{b} 2 \pi x \ell(x) d x \)

Step by step solution

01

Understand the problem

Given a planar region R with a length function 饾憴, we aim to find the volume of the 3D solid formed by revolving R around the y-axis. The volume of small disks formed by revolving a small length d饾懃 in the region R can be approximated as the volume of a cylinder. The volume of a cylinder is given by the formula "Height 脳 蟺 脳 (Radius)^2".
02

Setup the integral

We need to set up the volume of the solid by revolving about the y-axis as an integral. Since the volume of the cylinder-like disks formed by revolving d饾懃 is 蟺 脳 饾懃^2 脳 饾憴(饾懃), we can obtain the total volume by integrating this function over the interval [饾憥,饾憦]: $$ V = \int_{a}^{b} 2 \pi x \ell(x) d x $$ This is the integral we need to prove, based on the given information.
03

Proving the formula for a given planar region

To prove this formula, note that it is based on the fact that 饾憴(饾懃) is multiplied by \(2 \pi x\), representing the circumference of the cylinder. Thus, we need to show that the total length of the region R, gathered by revolving it around the y-axis, is properly transformed into volume through the integral. The lines given by 饾懃=饾憥 and 饾懃=饾憦 intersect the region R in line segments of total length 饾憴(饾憥) and 饾憴(饾憦), respectively. For every point on these line segments, the distance to the y-axis, along with the angle it makes with the y-axis, uniquely determines a point in the solid body formed by revolving R around the y-axis. Thus, the volume of the solid can be approximated by integrating the sum of the volumes of the cylindrical disks formed by revolving each of these line segments about the y-axis: $$ V \approx \sum_{s=a}^b 2 \pi s \ell(s) \Delta s $$ Since the function 饾憴:[饾憥,饾憦]鈫掆劃 is integrable, taking the limit as the size of the partition goes to zero gives us the desired result: $$ V = \lim_{\Delta s \to 0} \sum_{s=a}^b 2 \pi s \ell(s) \Delta s = \int_{a}^{b} 2 \pi x \ell(x) d x $$ This proves that the volume of the 3D solid formed by revolving R about the y-axis is equal to the given integral: $$ V = 2\pi\int_{a}^{b} x \ell(x) d x $$

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Key Concepts

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

Planar Region Volume
Calculating the volume of a three-dimensional shape generated by revolving a two-dimensional planar region around an axis can seem daunting, but calculus offers a straight-forward way to tackle it. The technique involves creating a series of thin, disk-like shapes (or washers) and summing their individual volumes to approximate the total volume of the solid.

Imagine slicing the planar region into vertical slices of infinitesimal width, dx. Each slice, when rotated around the y-axis, forms a cylindrical disk with radius x (the distance from the y-axis) and height equal to the length of the segment intersecting the region, which is \(\ell(x)\). The volume of each infinitesimally thin disk is then \(2\pi x \ell(x)dx\), where \(2\pi x\) is the circumference of the disk, effectively wrapping the height \(\ell(x)\) around the y-axis. By integrating this expression from a to b, we sum up all the infinitesimally small disks to find the entire volume of the solid of revolution.
Solid of Revolution

Creating a Solid by Rotation

If the function \( \ell(x)\) is integrable and describes the length of the line segments at a given x, the shape that results from revolving \( R\) about the y-axis is called a solid of revolution. Each value of x provides a different radius for the cylindrical disk, and the resulting three-dimensional object can have varying shapes depending on the nature of \( \ell(x)\).

The Washer Method

It's important to note that if \( R\) isn't a simple region but instead includes gaps, the cylindrical disks become washers鈥攔ings with inner and outer radii. The process still holds, but the volume calculation must account for these differences, involving the subtraction of the inner empty cylinder (if any) from the outer filled cylinder.

The integration process essentially stacks all these disks and washers along the y-axis, progressively adding up their volumes from \( x=a\) to \( x=b\). As such, we can visualize the solid as a collection of coins of varying sizes, closely packed to form a complete three-dimensional object.
Integrable Functions
For a function to be used in finding the volume of a solid of revolution, it is crucial that it must be integrable over the chosen interval. An integrable function is one that can be integrated鈥攊n other words, its integral exists within the interval in question. This property ensures that there is a well-defined area under the function's curve and, correspondingly, a well-defined volume can be calculated.

Suppose that \( \ell(x)\) represents the length of line segments at any x within the region \( R\) and is integrable. This means that we can take the sum of an infinite number of these lengths as x ranges from a to b, multiplied by their respective circumferences, to derive the total volume of the solid upon revolution. The function鈥檚 integrability is vital; without it, the volume calculation could lead to inconsistencies or even be impossible. Therefore, ensuring that \( \ell(x)\) is integrable is a prerequisite for applying the calculus techniques that enable us to find volumes by revolution.

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Most popular questions from this chapter

Let a curve \(C\) in \(\mathbb{R}^{2}\) be given by \((x(t), y(t)), t \in[\alpha, \beta] .\) For a partition \(\left\\{t_{0}, t_{1}, \ldots, t_{n}\right\\}\) of \([\alpha, \beta]\), let $$ \ell(C, P):=\sum_{i=1}^{n} \sqrt{\left(x\left(t_{i}\right)-x\left(t_{i-1}\right)\right)^{2}+\left(y\left(t_{i}\right)-y\left(t_{i-1}\right)\right)^{2}} . $$ If the set \(\\{\ell(C, P): P\) is a partition of \([\alpha, \beta]\\}\) is bounded above, then the curve \(C\) is said to be rectifiable, and the length of \(C\) is defined to be \(\ell(C):=\sup \\{\ell(C, P): P\) is a partition of \([\alpha, \beta]\\}\) (i) If \(\gamma \in(\alpha, \beta)\), and the curves \(C_{1}\) and \(C_{2}\) are given by \((x(t), y(t))\), \(t \in[\alpha, \gamma]\), and by \((x(t), y(t)), t \in[\gamma, \beta]\), respectively, then show that \(C\) is rectifiable if and only if \(C_{1}\) and \(C_{2}\) are rectifiable. (ii) Suppose that the functions \(x\) and \(y\) are differentiable on \([\alpha, \beta]\), and one of the derivatives \(x^{\prime}\) and \(y^{\prime}\) is continuous on \([\alpha, \beta]\), while the other is integrable on \([\alpha, \beta] .\) Show that the curve \(C\) is rectifiable and $$ \ell(C)=\int_{\alpha}^{\beta} \sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} d t $$

A twisted solid is generated as follows. A fixed line \(L\) in 3-space and a square of side \(s\) in a plane perpendicular to \(L\) are given. One vertex of the square is on \(L\). As this vertex moves a distance \(h\) along \(L\), the square turns through a full revolution with \(L\) as the axis. Find the volume of the solid generated by this motion. What would the volume be if the square had turned through two full revolutions in moving the same distance along the line \(L ?\)

Let \(\ell, \phi \in \mathbb{R}\) with \(\ell>0\). Consider the line segment given by \(\theta=\alpha(r)\), where \(\alpha(r):=\varphi\) for \(r \in[0, \ell]\). If this line segment is revolved about the \(x\) -axis, show that the area of the cone \(S\) so generated is equal to \(\pi \ell^{2}|\sin \varphi|\). [Note: Since the right circular cone \(S\) has slant height \(\ell\) and base radius \(\ell|\sin \varphi|\), the result matches the earlier calculation of the surface area of a right circular cone done by splitting it open.]

Let \(a \in \mathbb{R}\). Define \(f(x):=x-x^{2}\) and \(g(x):=a x\) for \(x \in \mathbb{R} .\) Determine \(a\) such that the region above the graph of \(g\) and below the graph of \(f\) has area equal to \(\frac{9}{2}\).

Show that the area of the elliptical region given by \(a x^{2}+2 b x y+c y^{2} \leq 1\), where \(a, b, c \in \mathbb{R}, c>0\), and \(a c-b^{2}>0\), is equal to \(\pi / \sqrt{a c-b^{2}}\).
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