Mathematics, 04.04.2020 19:47, HELPPPPPfawfds

Find the volume of a solid generated by revolving the region bounded by the graphs of the equations about the y-axis. Y= 4(3-x), Y= 0, and X= 0.

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answered: ottie123

$\displaystyle V = 36 \pi$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality

Algebra I

GraphingCoordinates (x, y)FunctionsFunction NotationIntersection PointsExpand by FOIL

Calculus

Integrals

Area under the curve

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

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

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

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

Volume of Revolution Formula [y-axis]: $\displaystyle V = \pi \int\limits^b_a {r^2} \, dy$

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: $\displaystyle \frac{y}{4} = 3 - x$ [Subtraction Property of Equality] Subtract 3 on both sides: $\displaystyle \frac{y}{4} - 3 = -x$ [Division Property of Equality] Divide -1 on both sides: $\displaystyle 3 - \frac{y}{4} = x$ Rewrite: $\displaystyle x = 3 - \frac{y}{4}$

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]: $\displaystyle V = \pi \int\limits^{12}_0 {(3 - \frac{y}{4})^2} \, dy$ [Integrand] Expand [FOIL]: $\displaystyle V = \pi \int\limits^{12}_0 {(\frac{y^2}{16} - \frac{3y}{2} + 9)} \, dy$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle V = \pi \bigg[ \int\limits^{12}_0 {\frac{y^2}{16}} \, dy - \int\limits^{12}_0 {\frac{3y}{2}} \, dy + \int\limits^{12}_0 {9} \, dy \bigg]$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle V = \pi \bigg[ \frac{1}{16} \int\limits^{12}_0 {y^2} \, dy - \frac{3}{2} \int\limits^{12}_0 {y} \, dy + 9 \int\limits^{12}_0 {} \, dy \bigg]$ [Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle V = \pi \bigg[ \frac{1}{16}(\frac{y^3}{3}) \bigg| \limits^{12}_0 - \frac{3}{2}(\frac{y^2}{2}) \bigg| \limits^{12}_0 + 9(y) \bigg| \limits^{12}_0 \bigg]$ [Integrals] Evaluate [Integration Rule - FTC 1]: $\displaystyle V = \pi \bigg[ \frac{1}{16}(576) - \frac{3}{2}(72) + 9(12) \bigg]$ [Brackets] Multiply: $\displaystyle V = \pi [36 - 108 + 108]$ [Brackets] Add: $\displaystyle V = \pi [36]$ Multiply: $\displaystyle V = 36 \pi$

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

Unit: Applications of Integration

Book: College Calculus 10e