(b-2)² + 10 for b = 7

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

harina [27]

Answer:

The volume of the solid is 714.887 units³

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: V = $\int\limits^a_b {2\pi xf(x)} \, dx$ ,

where 2πx · f(x) is the surface area of the cylinder shell and

dx is its thickness

* Lets solve the problem

∵ y = $x^{\frac{5}{2}}$

∵ The plane region is revolving about the y-axis

∵ y = 32 and x = 0

- Lets find the volume by the shell method

- The definite integral are x = 0 and the value of x when y = 32

- Lets find the value of x when y = 0

∵ $y = x^{\frac{5}{2}}$

∵ y = 32

∴ $32=x^{\frac{5}{2}}$

- We will use this rule to find x, if $x^{\frac{a}{b}}=c, then=== x=c^{\frac{b}{a}}$ , where c

is a constant

∴ $x=(32)^{\frac{2}{5}}=4$

∴ The definite integral are x = 0 , x = 4

- Now we will use the rule

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

∵ y = f(x) = x^(5/2) , a = 4 , b = 0

∴ $V=2\pi \int\limits^4_0 {x}.x^{\frac{5}{2}}\, dx$

- simplify x(x^5/2) by adding their power

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{7}{2}}} \, dx$

- The rule of integration of $x^{n} is ==== \frac{x^{n+1}}{(n+1)}$

∴ $V = 2\pi \int\limits^4_0 {x^{\frac{9}{2}}} \, dx=2\pi[\frac{x^{\frac{9}{2}}}{\frac{9}{2}}]$ from x = 0 to x = 4

∴ $V=2\pi[\frac{2}{9}x^{\frac{9}{2}}]$ from x = 0 to x = 4

- Substitute x = 4 and x = 0

∴ $V=2\pi[\frac{2}{9}(4)^{\frac{9}{2}}-\frac{2}{9}(0)^{\frac{9}{2}}}]=2\pi[\frac{1024}{9}-0]$

∴ $V=\frac{2048}{9}\pi=714.887$

* The volume of the solid is 714.887 units³

5 0

3 years ago

What is the solution? y = 3x - 5 <br> y = 1\3*+3

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

y = 3x - 5

y = 1/3x + 3

3x - 5 = 1/3x + 3....multiply everything by 3 to get rid of the fraction

9x - 15 = x + 9

9x - x = 9 + 15

8x = 24

x = 24/8

x = 3

y = 3x - 5

y = 3(3) - 5

y = 9 - 5

y = 4

solution is : x = 3 and y = 4...or (3,4) <==

3 0

3 years ago

A package of 30 Oreos cost $3.60. Kennedy says the unit rate is 0.12. Mason says the unit rate is 8.33. Who is right?

Alekssandra [29.7K]

Answer:

$360 \div 30 = 12$

Step-by-step explanation:

so Kennedy is right

8 0

2 years ago

At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee

masha68 [24]

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

5 0

3 years ago

Help me pleaseeeeeeee

puteri [66]

It would be the 30/3 one

7 0

3 years ago

(b-2)² + 10 for b = 7​

(b-2)² + 10 for b = 7