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Determining volumes II

Of course I’m in shape. Round is a shape.

Recall

Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.

Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.

The process of finding antiderivatives is called integration.

The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$

Calculating volumes with integrals is tricky and involves a variety of techniques such as slicing, the shell method, and the washer method.

Solved examples

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This is quite simple, A =[Each cross section is a semicircle] $\frac{1}{2}πr^2 =$ [d = 2sin(x) ⇒ r = sin(x)] $\frac{1}{2}π(sin^2x)$.

V = $\int_{0}^{π} \frac{1}{2}π(sin^2x)dx = \frac{1}{2}π\int_{0}^{π}(sin^2x) =\frac{1}{2}π\int_{0}^{π}\frac{1}{2}(1-cos(2x))dx = \frac{1}{4}π·(\int_{0}^{π} dx - \int_{0}^{π} cos(2x)dx) = \frac{1}{4}π·(x -\frac{1}{2}sin(2x))\bigg|_{0}^{π} = \frac{1}{4}π(π-\frac{1}{2})sin(2π)-(0-\frac{1}{2}sin(0)) = \frac{1}{4}π^2 ≈ 2.467 cm^3$

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The area of an equilateral triangle can be calculated using the formula: A(x) = $\frac{\sqrt{3}}{4}s^2$ where A is the area and s is the length of one side of the equilateral triangle. Since $x^2+y^2=4 ⇒ y^2=4-x^2$ ⇒[First quadrant] $y = \sqrt{4-x^2} ⇒ S = 2\sqrt{4-x^2} ⇒ A(x) = \frac{\sqrt{3}}{4}4(4-x^2) = \sqrt{3}(4-x^2)$

V =[the circle x2 + y2 = 4 has a radius r = 2 and x ∈ [-2, 2]] $\int_{-2}^{2} \sqrt{3}(4-x^2)dx = \sqrt{3}(4x -\frac{1}{3}x^3)\bigg|_{-2}^{2} = \sqrt{3}(8 -\frac{1}{3}8-(-8-\frac{1}{3}(-8))) = \sqrt{3}(8 -\frac{1}{3}8+8+\frac{1}{3}(-8)) = \sqrt{3}(16-\frac{16}{3}) = \sqrt{3}(\frac{48}{3}-\frac{16}{3}) = \frac{32·\sqrt{3}}{3}$

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The volume of a solid that extends from y = a to y = b and has a known integrable cross-sectional area A(y) perpendicular to the x-axis is given by the formula for the general slicing method, that is, V = $\int_{a}^{b} A(y)dx = \int_{0}^{2} A(y)dy$.

The equation of the line passing through two points (x1, y1) and (x2, y2) is $y -y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)$ ⇒ The line that crosses between (0, 2) and (2, 0) is $y-2=\frac{-2}{2}(x-0) ↭ y - 2 = -x$ ↭ y = 2 - x.

A(y) =[Formula for the area of a semicircle] $\frac{π}{2}r^2$ where r is the radius =[Since the diameter is 2-x, the radius of the semicircle is half that quantity, but we need it as a function of y. y = 2-x ↭ x = 2-y] $\frac{π}{2}*(\frac{2-y}{2})^2$

V = $\int_{0}^{2} A(y)dy = \int_{0}^{2} \frac{π}{2}*(\frac{2-y}{2})^2dy = \frac{π}{8} \int_{0}^{2} (4-4y+y^2)dy = \frac{π}{8}(4y+\frac{-4}{2}y^2+\frac{y^3}{3}) = \frac{π}{8}(4y-2y^2+\frac{y^3}{3}) = \bigg|_{0}^{2} = \frac{π}{8}(8-8+\frac{8}{3}) = \frac{π}{3}.$

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  1. x-intersects: 4 -x2 = 0, x^2 = 4, x = ±2.
  2. V = $\int_{a}^{b} A(y)dy = \int_{0}^{4} A(y)dy.$
  3. Area = $\frac{1}{2}πr^2$ where r = 4 - x2 ⇒[However, we want r = r(y), y = 4 -x2 ↭ x2 = 4 -y] r = $\sqrt{4 -y}, Area = \frac{1}{2}π\sqrt{4 -y}^2 = \frac{1}{2}π(4-y)$
  4. V = $\int_{0}^{4} A(y)dy = \int_{0}^{4} \frac{1}{2}π(4-y)dy = \frac{1}{2}π(4y -\frac{1}{2}y^2)\bigg|_{0}^{4} = \frac{1}{2}π(16-8)=4π$

A cone is a three-dimensional solid geometric shape having a circular base and a pointed edge at the top called the apex. A right circular cone is a cone where the axis of the cone is the line meeting the vertex to the midpoint of the circular base (Figure ii).

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We are going to integrate along the height of the cone and choose x to be the vertical direction. The cross sections are simply circles, but the radius varies from the base of the cone to the apex.

By similarity of triangles, $\frac{r}{h-x} = \frac{R}{h} ⇒ r = \frac{R}{h}(h-x) ⇒ A(x) = π\frac{R^2}{h^2}(h-x)^2, V = \int_{0}^{h} π\frac{R^2}{h^2}(h-x)^2dx = π\frac{R^2}{h^2}\int_{0}^{h}(h-x)^2dx = $[By substitution of variables u = h -x, du = -dx, u1 =h, u2 = 0] $π\frac{R^2}{h^2}(-\int_{h}^{0}u^2du) = π\frac{R^2}{h^2}\frac{-u^3}{3}\bigg|_{h}^{0} = π\frac{R^2}{h^2}(-0+\frac{h^3}{3}) = \frac{π}{3}R^2h$

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The solid of revolution is formed by revolving the region R around the x-axis ⇒ the cross-sections are circles, A(x) = πr2 where r = f(x)

V = $\int_{0}^{ln(4)} π(e^{-x})^2dx = \int_{0}^{ln(4)} π(e^{-2x})dx = π\frac{-1}{2}e^{-2x}\bigg|_{0}^{ln(4)} = \frac{-π}{2}(e^{-2ln(4)}-e^0) = \frac{-π}{2}(e^{ln(4)^{-2}}-1) = \frac{-π}{2}(\frac{1}{16}-1) = \frac{-π}{2}·\frac{-15}{16} = \frac{15π}{32}$

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The solid of revolution is formed by revolving the region around the x-axis ⇒ the cross-sections are circles, A(x) = πr2 where r = f(x)

V = $\int_{0}^{\frac{1}{2}} π(\frac{1}{\sqrt[4]{1-x^2}})^2dx = π\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}}dx = π·sin^{-1}(x)\bigg|_{0}^{\frac{1}{2}} = π·(sin^{-1}(\frac{1}{2})-sin^{-1}(0)) = π(\frac{π}{6}-0) = \frac{π^2}{6}.$

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The solid of revolution is formed by revolving the region around the x-axis ⇒ the cross-sections are circles, A(x) = πr2 where r = f(x) = x2-4x +5, and the volume is $\int_{1}^{4} πr^2dx = \int_{1}^{4} π(x^2-4x+5)^2dx = π\int_{1}^{4}(x^4-8x^3+26x^2-40x+25) = π(\frac{x^5}{5}-8\frac{x^4}{4}+26\frac{x^3}{3}-40\frac{x^2}{2}+25x) = π(\frac{x^5}{5}-2x^4+26\frac{x^3}{3}-20x^2+25x)\bigg|_{1}^{4} ≈ 49.009$

Let f be continuous and non-negative, let R be the region bounded above by the graph of x, below by the x-axis, on the left by x = a, and on the right by x = b. Then, the volume of the solid of revolution formed by revolving R around the x-axis is given by V = $\int_{a}^{b} πf(x)^2dx$

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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