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Use cylindrical shells to find the volume of the solid generated when the region
R under y = x2 over the interval (0,2) revolved about the line y = -1

Use cylindrical shells to find the volume of the solid generated when the region R under y x2 over the interval 02 revolved about the line y 1 class=

Respuesta :

Space

Answer:

[tex]\displaystyle V = \frac{176 \pi}{15}[/tex]

General Formulas and Concepts:

Pre-Algebra

  • Equality Properties

Algebra I

  • Terms/Coefficients
  • Expanding
  • Functions
  • Function Notation
  • Graphing
  • Exponential Rule [Root Rewrite]:                                                                     [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

Calculus

Integrals

  • Definite Integrals
  • Area under the curve

Integration Rule [Reverse Power Rule]:                                                                  [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:                                        [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:                                                              [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                           [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Shell Method:

[tex]\displaystyle V = 2\pi \int\limits^b_a {xf(x)} \, dx[/tex]

  • [Shell Method] x is the radius
  • [Shell Method] 2πx is the circumference
  • [Shell Method] 2πxf(x) is the surface area
  • [Shell Method] 2πxf(x)dx is the volume

Step-by-step explanation:

Step 1: Define

Identify

Graph of region

y = x²

x = 2

y = 4

Axis of Revolution: y = -1

Step 2: Sort

We are revolving around a horizontal line.

  1. [Function] Rewrite in terms of y:                                                                      x = √y
  2. [Graph] Identify bounds of integration:                                                       ��   [0, 4]

Step 3: Find Volume Pt. 1

  1. [Shell Method] Find distance of radius x:                                                       [tex]x = y + 1[/tex]
  2. [Shell Method] Find circumference variable f(x) [Area]:                                 [tex]\displaystyle f(x) = 2 - \sqrt{y}[/tex]
  3. [Shell Method] Substitute in variables:                                                           [tex]\displaystyle V = 2\pi \int\limits^4_0 {(y + 1)(2 - \sqrt{y})} \, dy[/tex]
  4. [Integral] Rewrite integrand [Exponential Rule - Root Rewrite]:                    [tex]\displaystyle V = 2\pi \int\limits^4_0 {(y + 1)(2 - y^\bigg{\frac{1}{2}})} \, dy[/tex]
  5. [Integral] Expand integrand:                                                                            [tex]\displaystyle V = 2\pi \int\limits^4_0 {(-y^\bigg{\frac{3}{2}} + 2y - y^\bigg{\frac{1}{2}} + 2)} \, dy[/tex]
  6. [Integral] Integrate [Integration Rule - Reverse Power Rule]:                        [tex]\displaystyle V = 2\pi \bigg( \frac{-2y^\bigg{\frac{5}{2}}}{5} + y^2 - \frac{2y^\bigg{\frac{3}{2}}}{3} + 2y \bigg) \bigg| \limits^4_0[/tex]
  7. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:              [tex]\displaystyle V = 2\pi (\frac{88}{15})[/tex]
  8. Multiply:                                                                                                             [tex]\displaystyle V = \frac{176 \pi}{15}[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Applications of Integration

Book: College Calculus 10e

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