Answer:
[tex]\displaystyle V = 36 \pi[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
Algebra I
- Graphing
- Coordinates (x, y)
- Functions
- Function Notation
- Intersection Points
- Expand by FOIL
Calculus
Integrals
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]
Volume of Revolution Formula [y-axis]: [tex]\displaystyle V = \pi \int\limits^b_a {r^2} \, dy[/tex]
Step-by-step explanation:
Step 1: Define
Identify
y = 4(3 - x)
y = 0
x = 0
Step 2: Redefine
Rewrite (Revolving around y-axis)
- [Division Property of Equality] Divide 4 on both sides: [tex]\displaystyle \frac{y}{4} = 3 - x[/tex]
- [Subtraction Property of Equality] Subtract 3 on both sides: [tex]\displaystyle \frac{y}{4} - 3 = -x[/tex]
- [Division Property of Equality] Divide -1 on both sides: [tex]\displaystyle 3 - \frac{y}{4} = x[/tex]
- Rewrite: [tex]\displaystyle x = 3 - \frac{y}{4}[/tex]
Step 2: Find Bounds of Integration
See attachment
Look at y-values, right to left.
Bounds: [0, 12]
Step 3: Find Volume
- Substitute in variables [Volume of Revolution Formula]: [tex]\displaystyle V = \pi \int\limits^{12}_0 {(3 - \frac{y}{4})^2} \, dy[/tex]
- [Integrand] Expand [FOIL]: [tex]\displaystyle V = \pi \int\limits^{12}_0 {(\frac{y^2}{16} - \frac{3y}{2} + 9)} \, dy[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle V = \pi \bigg[ \int\limits^{12}_0 {\frac{y^2}{16}} \, dy - \int\limits^{12}_0 {\frac{3y}{2}} \, dy + \int\limits^{12}_0 {9} \, dy \bigg][/tex]
- [Integrals] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle V = \pi \bigg[ \frac{1}{16} \int\limits^{12}_0 {y^2} \, dy - \frac{3}{2} \int\limits^{12}_0 {y} \, dy + 9 \int\limits^{12}_0 {} \, dy \bigg][/tex]
- [Integrals] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle V = \pi \bigg[ \frac{1}{16}(\frac{y^3}{3}) \bigg| \limits^{12}_0 - \frac{3}{2}(\frac{y^2}{2}) \bigg| \limits^{12}_0 + 9(y) \bigg| \limits^{12}_0 \bigg][/tex]
- [Integrals] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle V = \pi \bigg[ \frac{1}{16}(576) - \frac{3}{2}(72) + 9(12) \bigg][/tex]
- [Brackets] Multiply: [tex]\displaystyle V = \pi [36 - 108 + 108][/tex]
- [Brackets] Add: [tex]\displaystyle V = \pi [36][/tex]
- Multiply: [tex]\displaystyle V = 36 \pi[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Applications of Integration
Book: College Calculus 10e
