Revolution of Object Questions

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Mathematics

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Q2 Lumen OHM Х + u Х ohm.lumenlearning.com/assess2/?cid=44051&aid=3200642#/skip/5 B Question 5 < > B0/1 pt 2 397 Details Let C be the curve y = 5x3 for 0 < x < 1.2. 8 7 6 5 3 2 1 0.5 Find the surface area of revolution of C about the x-axis. Surface area = Question Help: D Video Submit Question Type here to search O TT . 9:37 AM 5/24/2020 1 2 Lumen OHM Х Ex: Determine the Center of Mass X + O Х ohm.lumenlearning.com/assess2/?cid=44051&aid=3200644#/skip/3 ES B lumenohm Jasraj Singh & User Settings My Classes Log Out online homework manager Course Messages Forums Calendar Gradebook Home > MATH 191 > Assessment 5.4 Homework VO Score: 4/5 3/5 answered Question 3 < > 0.5/1 pt 5297 Details Find the centroid (ī, T) of the region bounded by: y = 2.c+ 30, y = 0, x = 0, and x=8 c= y= Submit Question Type here to search TT O IT . 9:53 AM 5/24/2020 2 2 Lumen OHM Х Ex: Determine the Center of Mass X + O Х ohm.lumenlearning.com/assess2/?cid=44051&aid=3200644#/skip/5 ES B : lumenohm Jasraj Singh & User Settings My Classes Log Out online homework manager Course Messages Forums Calendar Gradebook Home > MATH 191 > Assessment 5.4 Homework 6 VO Score: 4/5 3/5 answered Question 5 < > 50.5/1 pt 92 98 Details Find the centroid of the region bounded by the graphs of the functions 1 T y = 5 sin(x), y = and touching the origin. 2 이 Ā®, and x = The centroid is at (ū, y) where c = Question Help: D Video Submit Question Type here to search TT O IT . 9:53 AM 5/24/2020 2
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Explanation & Answer

Attached.

Running Head: SURFACE AREA

1

Revolution of object
First Middle Last
name of Institution

SURFACE AREA

2

Question 5
Curve C 𝑦 = 5𝑥 3 ; 0 ≤ 𝑥 ≤ 1.2
Required: surface area of revolution of C about x-axis.
𝑑𝑦 2

𝑥=𝑏

Surface area 𝐴 = ∫ 2𝜋𝑦𝑑𝐿 = ∫𝑥=𝑎 2𝜋𝑦 √1 + (𝑑𝑥 ) 𝑑𝑥
𝑦 = 5𝑥 3
𝑑𝑦
= 15𝑥 2
𝑑𝑥
𝑑𝑦 2
( ) = 225𝑥 4
𝑑𝑥
Substituting back;
𝑏

𝐴 = ∫ 2𝜋𝑦 √1 + 225𝑥 4 𝑑𝑥
𝑎

Let
1 + 225𝑥 4 = 𝑡 2
Differentiating both sides;
2𝑡 𝑑𝑡 = 900𝑥 3 𝑑𝑥

𝑥 3 𝑑𝑥 =

2𝑡𝑑𝑡
𝑡𝑑𝑡
=
900
450

When 𝑥 = 0, 𝑡 = √1 + 225(0) = 1
When 𝑥 = 1.2, 𝑡 = √1 + 225(1.24 ) = 21.6231

SURFACE AREA

3

We now have
21.6231

21.6231

2𝜋(5𝑥 3 )√𝑡 2 𝑑𝑥

𝐴= ∫

= 10𝜋 ∫

1

𝑥 3 𝑡 𝑑𝑥

1

Remember
𝑥 3 𝑑𝑥 =

𝑡𝑑𝑡
450

Thus,

𝑑𝑥 =

𝑡𝑑𝑡
450𝑥 3

Substituting back in formula;
21.6231

𝐴 = 10𝜋 ∫
1

21.6231

𝑡𝑑𝑡
𝜋 21.6231 2
𝜋 𝑡3
𝑥 ∗𝑡∗
=

𝑡
𝑑𝑡
=
[ ]
450𝑥 3
45 1
45 3 1
3

𝐴=

𝜋 21.62313 13
[
− ] = 74.882𝜋
45
3
3

The surface area of revolution of C about x-axis is 74.882𝜋
Question 3
𝑦 = 2𝑥 2 + 3𝑥; 𝑦 = 0, 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 8
Required: centroid (𝑥̅ ,...


Anonymous
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