Question

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

Answer 1

Answer:

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

General Formulas and Concepts:

Pre-Algebra

• Equality Properties

Algebra I

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

Calculus

Integrals

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

Shell Method:

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

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

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

Unit: Applications of Integration

Book: College Calculus 10e