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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 ,...