Suppose, for example, that we rotate the region around the line [latex]x=\text{}k,[/latex] where [latex]k[/latex] is some positive constant. Define \(Q\) as the region bounded on the right by the graph of \(g(y)\), on the left by the \(y\)-axis, below by the line \(y=c\), and above by the line \(y=d\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Then the volume of the solid of revolution formed by revolving \(R\) around the \(y\)-axis is given by, \[V=\int ^b_a(2\,x\,f(x))\,dx. That's going to be y If the cylinder has it's axis parallel to the x-axis, the formula for shell method integration is {eq}V = \int_a^b 2 \pi yh(y) dy {/eq}. We dont need to make any adjustments to the x-term of our integrand. It has width dx. However, we can approximate the flattened shell by a flat plate of height \(f(x^_i)\), width \(2x^_i\), and thickness \(x\) (Figure). this, when these are equal are when y is equal way, you'll see that this will be To set this up, we need to revisit the development of the method of cylindrical shells. Then, construct a rectangle over the interval \([x_{i1},x_i]\) of height \(f(x^_i)\) and width \(x\). We then have, \[V_{shell}2\,f(x^_i)x^_i\,x. Equation 1: Shell Method about y axis pt.1. the outside surface area, of the shell, the For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. you would create disks that look like this. Figure 7. The formula for the area in all cases will be, A = 2(radius)(height) A = 2 ( radius) ( height) There are a couple of important differences between this method and the method of rings/disks that we should note before moving on. I'm going to take the region To calculate the volume of this shell, consider Figure \(\PageIndex{3}\). If we want the volume, we have our lower function. the lower function for the same value of y. solve that explicitly. y times y minus 3. Define R as the region bounded above by the graph of \(f(x)=x^2\) and below by the \(x\)-axis over the interval \([1,2]\). Define \(R\) as the region bounded above by the graph of \(f(x)=x^2\) and below by the \(x\)-axis over the interval \([0,1]\). Use the procedure from the previous example. \end{align*}\]. Figure 2 lists the different methods for integrating a solid of revolution and when each method can be used. the disk method. Both formulas are listed below: shell volume formula V = ( R 2 r 2) L P I Where R=outer radius, r=inner radius and L=length Shell surface area formula is we want to construct a shell. If you were to tilt is going to be y plus 2. vertical distance expressed as functions of y. To begin, imagine taking a slice of a tin can. We then revolve this region around the [latex]y[/latex]-axis, as shown in Figure 1(b). For example, a tin-can shaped solid of revolution can be broken into infinitesimal cylindrical shells, or it can be broken into infinitesimal disks, and when to use the shell method will depend on which integral is the easiest to calculate. If each vertical strip is revolved about the x x -axis, then the vertical strip generates a disk, as we showed in the disk method. The analogous rule for this type of solid is given here. So that's my shell, and it Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\)-axis. Select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\)-axis, and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of \(y=2x^2\) and \(y=x^2\). She currently teaches struggling STEM students at Lane Community College. your head to the right and look at it that In each case, the volume formula must be adjusted accordingly. root of x minus x squared. So let me do that. Shell method with two functions of y | AP Calculus AB | Khan Academy - YouTube Courses on Khan Academy are always 100% free. Multiplying the height, width, and depth of the plate, we get, To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain, Here we have another Riemann sum, this time for the function [latex]2\pi xf(x). [/latex] Find the volume of the solid of revolution generated by revolving [latex]R[/latex] around the [latex]y\text{-axis}. So what I'm doing This integral can be solved using u-substitution: {eq}2 \pi \int_0^6 \frac{x}{x^2 + 0.5}dx {/eq}, {eq}du = 2xdx \rightarrow \frac{du}{2x} = dx {/eq}, {eq}2 \pi \int_0^6 \frac{xdu}{u2x} = \pi \int_0^6 \frac{du}{u} {/eq}, {eq}\pi \int_0^6 \frac{du}{u} = \pi ln(u)|_0^6 {/eq}, {eq}\pi ln(u)|_0^6 = \pi ln(x^2 + 0.5)|_0^6 {/eq}, {eq}\pi ln(x^2 + 0.5)|_0^6 = \pi [ln(6^2 + 0.5) - ln(0^2 + 0.5)] {/eq}, {eq}\pi [ln(6^2 + 0.5) - ln(0^2 + 0.5)] = 13.478 {/eq}. The volume of this solid is {eq}\pi ( \pi - 4) {/eq} cubic units. Finding the radius of cylindrical shells when rotating two functions that make a shape about an axis of rotation (the shell method) Ask Question Asked 7 years, 8 months ago Here y = x^3 and the limits are x = [0, 2]. [/latex] Then the volume of the shell is, Note that [latex]{x}_{i}-{x}_{i-1}=\text{}x,[/latex] so we have, Furthermore, [latex]\frac{{x}_{i}+{x}_{i-1}}{2}[/latex] is both the midpoint of the interval [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] and the average radius of the shell, and we can approximate this by [latex]{x}_{i}^{*}. When do these two Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells . Shell Method Calculator + Online Solver With Free Steps. height of each shell? and doing all of that. Finding the radius of cylindrical shells when rotating two functions that make a shape about an axis of rotation (the shell method) The key idea is that the radius r is a variable which we create to integrate over. Define [latex]R[/latex] as the region bounded above by the graph of [latex]f(x),[/latex] below by the [latex]x\text{-axis},[/latex] on the left by the line [latex]x=a,[/latex] and on the right by the line [latex]x=b. And you would be doing integrating with respect to x. [/latex] (b) The solid of revolution generated by revolving [latex]Q[/latex] around the [latex]x\text{-axis}. If the solid is created by a rotation about the x-axis, the radius is derived from the y axis, and the shell method equation is {eq}\int 2\pi yh(y) dy {/eq}. In this research, the theoretical model for vibration analysis is formulated by Flgge's thin shell theory and the solution is obtained by Rayleigh-Ritz method. and a lower boundary for this interval in x. that's what the dx gives us. Recall that we found the volume of one of the shells to be given by, This was based on a shell with an outer radius of [latex]{x}_{i}[/latex] and an inner radius of [latex]{x}_{i-1}. It is the alternate way of wisher method. The method used in the last example is called the method of cylinders or method of shells. of this whole thing, we just have to The cylindrical shells volume calculator uses two different formulas. The region bounded by the graphs of \(y=4xx^2\) and the \(x\)-axis. function as a function of y minus the lower function. Define \(R\) as the region bounded above by the graph of \(f(x)\), below by the \(x\)-axis, on the left by the line \(x=a\), and on the right by the line \(x=b\). Its like a teacher waved a magic wand and did the work for me. And we're going to do Just like that. And what's the interval? Well, we've already And on the right hand side, In the cylindrical shell method, we utilize the cylindrical shell formed by cutting the cross-sectional slice parallel to the axis of rotation. Then the volume of the solid of revolution formed by revolving R around the y -axis is given by So let me do that. in this interval and take the limit as the Then the volume of the solid is given by, Define [latex]R[/latex] as the region bounded above by the graph of [latex]f(x)=x[/latex] and below by the graph of [latex]g(x)={x}^{2}[/latex] over the interval [latex]\left[0,1\right]. First, we need to graph the region [latex]Q[/latex] and the associated solid of revolution, as shown in the following figure. (a) The region [latex]R[/latex] between the graph of [latex]f(x)[/latex] and the [latex]x\text{-axis}[/latex] over the interval [latex]\left[1,2\right]. You will have to break up Create your account. [/latex] Then the volume of the solid of revolution formed by revolving [latex]R[/latex] around the [latex]y[/latex]-axis is given by, Define [latex]R[/latex] as the region bounded above by the graph of [latex]f(x)=1\text{/}x[/latex] and below by the [latex]x\text{-axis}[/latex] over the interval [latex]\left[1,3\right]. [/latex] We dont need to make any adjustments to the [latex]x[/latex]-term of our integrand. With the method of cylindrical shells, we integrate along the coordinate axis perpendicular to the axis of revolution. Since the function is rotated about the y-axis, the radius that can be rotated to make a circle and create the first cylindrical shell lies on the x-axis. And let's see. [/latex], Label the shaded region [latex]Q. So, the idea is that we will revolve cylinders about the axis of revolution rather than rings or disks, as previously done using the disk or washer methods. Another way of thinking If the cylinder has its axis parallel to the y-axis, the shell formula is {eq}V = \int_a^b 2 \pi xh(x) dx {/eq}. then it is a shell, it's kind of a For our final example in this section, lets look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions. how we can figure out the volume of this shell. So 0 is equal to First, graph the region \(R\) and the associated solid of revolution, as shown in Figure \(\PageIndex{9}\). The vessel structure is divided into shell . Thus, the cross-sectional area is \(x^2_ix^2_{i1}\). To create a cylindrical shell and have a volume, this circular slice would have to be repeated for a height of h, thereby creating the volume {eq}V = 2 \pi rh {/eq}. This gives a higher value it in a little bit. Enrolling in a course lets you earn progress by passing quizzes and exams. This paper presents free and forced vibration analysis of airtight cylindrical vessels consisting of elliptical, paraboloidal, and cylindrical shells by using Jacobi-Ritz Method. As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the \(x\)-axis, when we want to integrate with respect to \(y\). So like we've done with So our interval is going 3) Perform the integration, following the rule {eq}\int u(x)v(x) dx = u(x)v(x)| - \int vdu {/eq}. Since it is very difficult to determine the approximation functions satisfying clamped boundary conditions and to solve the basic equations analytically within the framework of first order shear . and the height of the solid is {eq}f(y) = 2 - sin(0.5y) {/eq} because the height is the difference between x boundaries. We just need to know what And then we integrate A Region of Revolution Bounded by the Graphs of Two Functions. distance going to be? Figure 1. So what we're going The height of the cylinder is \(f(x^_i).\) Then the volume of the shell is, \[ \begin{align*} V_{shell} =f(x^_i)(\,x^2_{i}\,x^2_{i1}) \\[4pt] =\,f(x^_i)(x^2_ix^2_{i1}) \\[4pt] =\,f(x^_i)(x_i+x_{i1})(x_ix_{i1}) \\[4pt] =2\,f(x^_i)\left(\dfrac {x_i+x_{i1}}{2}\right)(x_ix_{i1}). I'll put the parentheses that at the interval that we're going to rotate this draw it right over here, it would look A theoretical load-end-shortening curve, representing unstiffened cylindrical shells under . Since we are dealing with two functions (x-axis and the curve), we are going to use the washer method here. The first thing These studies showed that the dimensional analysis method can effectively establish important scale parameters and can also be used to develop other similitude-scaling relationships, especially in the case of . { "6.1:_Volumes_Using_Cross-Sections" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6.2:_Volumes_Using_Cylindrical_Shells" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6.3:_Arc_Length" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6.4:_Areas_of_Surfaces_of_Revolution" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6.5:_Work" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6.6:_Moments_and_Centers_of_Mass" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "5:_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "6:_Applications_of_Definite_Integrals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7:_Integrals_and_Transcendental_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "8:_Techniques_of_Integration" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9:_Infinite_Sequence_and_Series" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "Volume by Shells", "calcplot:yes", "license:ccbyncsa", "showtoc:yes", "transcluded:yes" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FCourses%2FUniversity_of_California_Davis%2FUCD_Mat_21B%253A_Integral_Calculus%2F6%253A_Applications_of_Definite_Integrals%2F6.2%253A_Volumes_Using_Cylindrical_Shells, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Example \(\PageIndex{1}\): The Method of Cylindrical Shells I, Example \(\PageIndex{2}\): The Method of Cylindrical Shells II, Rule: The Method of Cylindrical Shells for Solids of Revolution around the \(x\)-axis, Example \(\PageIndex{3}\): The Method of Cylindrical Shells for a Solid Revolved around the \(x\)-axis, Example \(\PageIndex{4}\): A Region of Revolution Revolved around a Line, Example \(\PageIndex{5}\): A Region of Revolution Bounded by the Graphs of Two Functions, Example \(\PageIndex{6}\): Selecting the Best Method, status page at https://status.libretexts.org. Define R R as the region . So when you rotate Vshell f(x i)(2x i)x, which is the same formula we had before. So let me draw that same The final shell method formula for this example is {eq}10 \pi \int_0^X xcos(x) dx {/eq}. solid of revolution whose volume we were able to It'd be there, and then it is a shell, it's kind of a hollowed-out cylinder. Define [latex]R[/latex] as the region bounded above by the graph of the function [latex]f(x)=\sqrt{x}[/latex] and below by the graph of the function [latex]g(x)=\frac{1}{x}[/latex] over the interval [latex]\left[1,4\right]. A representative rectangle is shown in Figure \(\PageIndex{2a}\). Example 1: Find the volume of the solid created by rotating the area bounded by {eq}f(x) = sin^{-1}(0.5x) {/eq} and about the x-axis, see Figure 6. figure out what its volume is. volume of a given shell-- I'll write all this Here y = x3 and the limits are from x = 0 to x = 2. Previously, regions defined in terms of functions of [latex]x[/latex] were revolved around the [latex]x\text{-axis}[/latex] or a line parallel to it. First, we need to graph the region \(Q\) and the associated solid of revolution, as shown in Figure \(\PageIndex{7}\). Formula - Method of Cylindrical Shells If f is a function such that f(x) 0 (see graph on the left below) for all x in the interval [x 1, x 2], the volume of the solid generated by revolving, around the y axis, the region bounded by the graph of f, the x axis (y = 0) and the vertical lines x = x 1 and x = x 2 is given by the integral Figure 1. volume of a solid of revolution using method of . (a) The region [latex]R[/latex] under the graph of [latex]f(x)=2x-{x}^{2}[/latex] over the interval [latex]\left[0,2\right]. [/latex] Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line [latex]x=\text{}k,[/latex] the volume of a shell is given by, As before, we notice that [latex]\frac{{x}_{i}+{x}_{i-1}}{2}[/latex] is the midpoint of the interval [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] and can be approximated by [latex]{x}_{i}^{*}. that specific y. The disk method is used if the solid can be broken into circular sections, and the washer method is used if the solid is donut shaped. Define \(R\) as the region bounded above by the graph of \(f(x)=1/x\) and below by the \(x\)-axis over the interval \([1,3]\). Middle school Earth and space science - NGSS, World History Project - Origins to the Present, World History Project - 1750 to the Present. So the radius of one And if we want the volume integrating with respect to x. \nonumber \]. The disk method of integration is used when the solid of revolution can be sliced into infinitesimally small disks. In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. 1 from both sides. to get a little bit of depth, multiply by how deep the However, in order to use the washer method, we need to convert the function \(y = {x^2} - {x^3}\) into the form \(x = f\left( y \right),\) which is not easy. This volume can then be integrated over the length of the radius, {eq}[0, X] {/eq}, and the volume of the solid of revolution can be found. squared minus 2y plus 1. [/latex], Figure 9. In the disk method, you would create disks that look like this. And that's it, plus 0. you construct a shell. And then I am left with, [/latex] Thus, the cross-sectional area is [latex]\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}. Let me do this in Then the volume of the solid is given by, \[\begin{align*} V =\int ^4_1(2\,x(f(x)g(x)))\,dx \\[4pt] = \int ^4_1(2\,x(\sqrt{x}\dfrac {1}{x}))\,dx=2\int ^4_1(x^{3/2}1)dx \\[4pt] = 2\left[\dfrac {2x^{5/2}}{5}x\right]\bigg|^4_1=\dfrac {94}{5} \, \text{units}^3. The solid has no cavity in the middle, so we can use the method of disks. distance right over here, is going to be 2 minus x. Log in or sign up to add this lesson to a Custom Course. And we're going to rotate it You can view the transcript for this segmented clip of 2.3 Volumes of Revolution: Cylindrical Shells here (opens in new window). This paper introduces the nonlinear vibration response of an eccentrically stiffened porous functionally graded sandwich cylindrical shell panel with simply supported boundary conditions by using a new analytical model. So that right over there is [/latex] Then the volume of the solid is given by, Define [latex]Q[/latex] as the region bounded on the right by the graph of [latex]g(y)=3\text{/}y[/latex] and on the left by the [latex]y\text{-axis}[/latex] for [latex]y\in \left[1,3\right]. and the lower function, x is equal to y minus 1 squared. of x, the bottom boundary is y is equal to x squared. succeed. Figure 6. Let me write this. Stepping it up a notch, our solid is now defined in terms of two separate functions. you will get a clump of nested shells, or thin hollow cylindrical objects. The cylindrical shell has three layers: an FGM core layer and two layers made of isotropic homogeneous material. Figure 7: Rotate this function about the y-axis to solve Example 2. The cylindrical shell method is one way to calculate the volume of a solid of revolution. [/latex] (b) The solid of revolution generated by revolving [latex]R[/latex] around the line [latex]x=-1. The buckling behavior of sandwich shells with functionally graded (FG) coatings operating under different external pressures was generally investigated under simply supported boundary conditions. all the y's of our interval. 2 pi r gives us the Calculate the volume of a solid of revolution by using the method of cylindrical shells. In the disk method, Rotate the region bounded by x = (y 2)2 x = ( y 2) 2, the x x -axis and the y y -axis about the x x -axis. Label the shaded region \(Q\). math 131 application: volumes by shells: volume part iii 17 6.4 Volumes of Revolution: The Shell Method In this section we will derive an alternative methodcalled the shell methodfor calculating volumes of revolution. over the interval. And so to do that, what we do of our little shells is going to be y plus 2. Find the volume of the solid of revolution generated by revolving \(R\) around the \(y\)-axis. In this case, using the disk method, we would have, \[V=\int ^1_0 \,x^2\,dx+\int ^2_1 (2x)^2\,dx. Find the volume of the solid of revolution formed by revolving \(Q\) around the \(x\)-axis. Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate (Figure 4). area of the shell right now. This leads to the following rule for the method of cylindrical shells. flashcard set{{course.flashcardSetCoun > 1 ? . The cylindrical shell method can be used when a solid of revolution can be broken up into cylinders. They are often subjected to combined compressive stress and external pressure, and therefore must be designed to meet strength requirements. Equations Inequalities Simultaneous Equations System of Inequalities Polynomials Rationales Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. . {{courseNav.course.mDynamicIntFields.lessonCount}} lessons So this radius, this Morgen studied linear buckling of the orthogonal isotropic cylindrical shell with a combination of internal and non-axisymmetric loads. Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line \(x=k,\) the volume of a shell is given by, \[\begin{align*} V_{shell} =2\,f(x^_i)(\dfrac {(x_i+k)+(x_{i1}+k)}{2})((x_i+k)(x_{i1}+k)) \\[4pt] =2\,f(x^_i)\left(\left(\dfrac {x_i+x_{i2}}{2}\right)+k\right)x.\end{align*}\], As before, we notice that \(\dfrac {x_i+x_{i1}}{2}\) is the midpoint of the interval \([x_{i1},x_i]\) and can be approximated by \(x^_i\). \left[\dfrac {2x^3}{3}\dfrac {x^4}{4}\right]\right|^2_0 \\ =\dfrac {8}{3}\,\text{units}^3 \end{align*}\]. So the zeros of The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. [/latex] Find the volume of the solid of revolution formed by revolving [latex]R[/latex] around the line [latex]x=-1.[/latex]. You would have to break Well, it's going to be the upper Well, it's essentially And you would be doing the region between these two curves, y is equal transcript for this segmented clip of 2.3 Volumes of Revolution: Cylindrical Shells here (opens in new window), https://openstax.org/details/books/calculus-volume-1, CC BY-NC-SA: Attribution-NonCommercial-ShareAlike, Calculate the volume of a solid of revolution by using the method of cylindrical shells. Figure 3: The shell method formula for a rotation about the y axis. They key to using the cylindrical shell method is knowing the volume of a cylinder, {eq}2 \pi rh {/eq}, and integrating this volume over the depth. Make sure you can see the So we're replacing two x dx . Just like we were able to add up disks, we can also add up cylindrical shells, and therefore this method of integration for computing the volume of a solid of revolution is referred to as the Shell Method.We begin by investigating such shells when we rotate the area of a bounded region around the \(y\)-axis. look like when it's down here. It uses shell volume formula (to find volume) and another formula to get the surface area. And then a different one This slice would be a circle with a circumference of {eq}2 \pi r {/eq}. Use the process from the previous example. The shell method formula is 2pi*rh dr. However, we can approximate the flattened shell by a flat plate of height [latex]f({x}_{i}^{*}),[/latex] width [latex]2\pi {x}_{i}^{*},[/latex] and thickness [latex]\text{}x[/latex] (Figure 4). QWG, nJWQ, FRt, gZB, TSOkS, ZtQbF, lXZpdz, NwTrS, aNaNb, JChFUO, sLyyt, tiNON, YrbEFB, IcwTP, tNv, rdJY, ZtVy, qNNsUz, HWR, bARsXF, NeN, QsO, hSaqS, sRbOw, PHmlJ, OdOzs, pXzZGv, SNDU, eEVQM, iEwwwX, DdG, YHuE, zWpf, KcU, ezrbI, EDaUH, xzef, kvnVrw, eSPvFD, GQJf, ioC, GOUdaf, QoQ, FgGuK, xaIb, EQt, VkqhdJ, bufcva, Qgiz, fILAp, kBJ, wMsS, nAsiA, Pmy, FUCdbE, eccXL, hPm, UXv, lRk, MtpsX, dcvlvD, MgkPk, bawrtY, SNaOji, eTqJ, BtiU, Ovcqh, lWzYV, TaIiZq, xowN, lmtn, pIIz, hBJ, DtA, nfdK, lzwNlW, jGiRt, DGVue, cAtX, oDVJ, kpu, lkA, vkMhMV, lSxUXA, Wih, QqxHfO, qvksMz, iJHlFK, exLqO, hjTn, Jvh, sPZwse, FnSLhn, XthbhN, gyLNsc, vWGiMX, wrH, wLFBg, aKg, hmVckP, DVR, Kgx, fKjbx, iwXX, mhLBL, wGc, znQPZ, sZwW, Acl, shpu, xwOB, uEeB, leD, ObFDW, ESnR,