Use cylindrical shells to find the volume of the solid generated when the region

1 answer:

6 0

Answer:

$\displaystyle V = \frac{176 \pi}{15}$

General Formulas and Concepts:

Pre-Algebra

Algebra I

Calculus

Integrals

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]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Shell Method:

$\displaystyle V = 2\pi \int\limits^b_a {xf(x)} \, dx$

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.

Step 3: Find Volume Pt. 1

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