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A light bulb is designed by revolving the graph of:

[tex]y = \frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}[/tex]

Over the interval 0 ≤ x ≤ 1/3 about the x-axis, where x and y are measured in feet. Find the surface area of the bulb and use the result to approximate the amount of glass needed to make the bulb. (Glass is 0.015 inch thick).

Respuesta :

Space

Answer:

[tex]\displaystyle 0.251327 \ in. \ of \ glass[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Algebra I

  • Terms/Coefficients
  • Expand by FOIL (First Outside Inside Last)
  • Factoring

Calculus

Differentiation

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Integration

  • Integration Property: [tex]\displaystyle \int\limits^a_b {cf(x)} \, dx = c \int\limits^a_b {f(x)} \, dx[/tex]
  • Fundamental Theorem of Calculus: [tex]\displaystyle \int\limits^a_b {f(x)} \, dx = F(b) - F(a)[/tex]
  • Area between Two Curves
  • Volumes of Revolution
  • Arc Length Formula: [tex]\displaystyle AL = \int\limits^a_b {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]
  • Surface Area Formula: [tex]\displaystyle SA = 2\pi \int\limits^a_b {f(x) \sqrt{1+ [f'(x)]^2}} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle y = \frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}\\Interval: [0, \frac{1}{3}][/tex]

Step 2: Differentiate

  1. Basic Power Rule:                    [tex]\displaystyle y' = \frac{1}{2} \cdot \frac{1}{3}x^{\frac{1}{2} - 1} - \frac{3}{2} \cdot x^{\frac{3}{2} - 1}[/tex]
  2. [Derivative] Simplify:                [tex]\displaystyle y' = \frac{1}{6}x^{\frac{-1}{2}} - \frac{3}{2}x^{\frac{1}{2}}[/tex]
  3. [Derivative] Simplify:                [tex]\displaystyle y' = \frac{1}{6\sqrt{x}} - \frac{3\sqrt{x}}{2}}[/tex]

Step 3: Integrate Pt. 1

  1. Substitute [Surface Area]:                                                                             [tex]\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{1+ [\frac{1}{6\sqrt{x}} - \frac{3\sqrt{x}}{2}}]^2}} \, dx[/tex]
  2. [Integral - √Radical] Expand/Add:                                                               [tex]\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{\frac{81x^2+18x+1}{36x}} \, dx[/tex]
  3. [Integral - √Radical] Factor:                                                                         [tex]\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {(\frac{1}{3}x^{\frac{1}{2}} - x^{\frac{3}{2}}) \sqrt{\frac{(9x + 1)^2}{36x}} \, dx[/tex]
  4. [Integral - Simplify]:                                                                                       [tex]\displaystyle SA = 2\pi \int\limits^{\frac{1}{3}}_0 {-\frac{|9x + 1|(3x - 1)}{18}} \, dx[/tex]
  5. [Integral] Integration Property:                                                                     [tex]\displaystyle SA = \frac{- \pi}{9} \int\limits^{\frac{1}{3}}_0 {|9x + 1|(3x - 1)} \, dx[/tex]

Step 4: Integrate Pt. 2

  1. [Integral] Define:                                                                                             [tex]\displaystyle \int {|9x + 1|(3x - 1)} \, dx[/tex]
  2. [Integral] Assumption of Positive/Correction Factors:                                 [tex]\displaystyle \frac{9x + 1}{|9x + 1|} \int {(9x + 1)(3x - 1)} \, dx[/tex]
  3. [Integral] Expand - FOIL:                                                                                 [tex]\displaystyle \frac{9x + 1}{|9x + 1|} \int {27x^2 - 6x - 1} \, dx[/tex]
  4. [Integral] Integrate - Basic Power Rule:                                                         [tex]\displaystyle \frac{9x + 1}{|9x + 1|} (9x^3 - 3x^2 - x)[/tex]
  5. [Expression] Multiply:                                                                                      [tex]\displaystyle \frac{(9x + 1)(9x^3 - 3x^2 - x)}{|9x + 1|}[/tex]

Step 5: Integrate Pt. 3

  1. [Integral] Substitute/Integral - FTC:                                                              [tex]\displaystyle SA = \frac{- \pi}{9} (\frac{(9x + 1)(9x^3 - 3x^2 - x)}{|9x + 1|})|\limits_{0}^{\frac{1}{3}}[/tex]
  2. [Integrate] Evaluate FTC:                                                                                [tex]\displaystyle SA = \frac{- \pi}{9} (\frac{-1}{3})[/tex]
  3. [Expression] Multiply:                                                                                     [tex]\displaystyle SA = \frac{\pi}{27} \ ft^2[/tex]

It is in ft² because it is given that our axis are in ft.

Step 6: Find Amount of Glass

Convert ft² to in² and multiply by 0.015 in (given) to find amount of glass.

  1. Convert ft² to in²:                    [tex]\displaystyle \frac{\pi}{27} \ ft^2 \ \div 144 \ in^2/ft^2 = \frac{16 \pi}{3} \ in^2[/tex]
  2. Multiply:                                   [tex]\displaystyle \frac{16 \pi}{3} \ in^2 \cdot 0.015 \ in = 0.251327 \ in. \ of \ glass[/tex]

And we have our final answer! Hope this helped on your Calc BC journey!

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