SOLUTION: Volume of A Sphere and Method of Cylindrical Shells Mathematics Questions

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Attached.

Solution
Intersection points:
𝑦 2 − 4𝑦 = 3𝑦 − 𝑦 2
𝑦 2 + 𝑦 2 − 4𝑦 − 3𝑦 = 0
2𝑦 2 − 7𝑦 = 0
𝑦(2𝑦 − 7) = 0 → 𝑦 = 0

𝑦(𝑦 − 4) = 𝑦(3 − 𝑦)
𝑦−4=3−𝑦
𝑦−4−3+𝑦 = 0

2𝑦 = 7 → 𝑦 =

7
2

𝑏

𝐴 = ∫[𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎
7
2

𝐴 = ∫ 3𝑦 − 𝑦 2 − (𝑦 2 − 4𝑦) 𝑑𝑦
0
7
2

= ∫ 3𝑦 − 𝑦 2 − 𝑦 2 + 4𝑦 𝑑𝑦
0
7
2

= ∫ −2𝑦 2 + 7𝑦 𝑑𝑦
0

7
2 3 7 2 2
= [− 𝑦 + 𝑦 ]
3
2
0
2 7 3 7 7 2
=− ( ) + ( ) −0
3 2
2 2

𝑨 = 𝟏𝟒. 𝟐𝟗

#2. Derive the quantity of a orb by revolving the semi-circle defined by
𝑦 = √𝑟 2 − 𝑥 2
circle.

Solution

about the x-axis where r is the (fixed) radius if the semi-

𝑥2

𝑟

2

𝑟

𝑉 = ∫ 𝜋𝑓(𝑥)2 𝑑𝑥 = ∫ 𝜋 (√𝑟 2 − 𝑥 2 ) 𝑑𝑥 = ∫ 𝜋(𝑟 2 − 𝑥 2 )𝑑𝑥
𝑥1

−𝑟

= 𝜋 [𝑟 2 𝑥 −

= 𝜋 [(𝑟 3 −

−𝑟

𝑥3 𝑟
]
3 −𝑟

𝑟3
𝑟3
) − (−𝑟 3 + )]
3
3

𝑽=

𝟒 𝟑
𝝅𝒓
𝟑

#3. Invent the quantity of the cap of the orb of radius r. The culmination of the
cap is h. (See delineate).

Solution
Disk Method
𝑏

𝑉 = ∫ 𝜋(𝑅𝑎𝑑𝑖𝑢𝑠)2 𝑑𝑦
𝑎
𝑟

𝑉 = ∫ 𝜋(𝑅𝑎𝑑𝑖𝑢𝑠)2 𝑑𝑦
𝑟−ℎ

𝑥 2 = 𝑟 2 − 𝑦 2 → 𝑥 = √𝑟 2 − 𝑦 2 → 𝑅𝑎𝑑𝑖𝑢𝑠

𝑟

𝑉 = 𝜋 ∫ 𝑟 2 − 𝑦 2 𝑑𝑦
𝑟−ℎ

1
𝑟
= 𝜋 (𝑟 2 𝑦 − 𝑦 3 )
3
𝑟−ℎ
1
1
= 𝜋 (𝑟 2 (𝑟) − 𝑟 3 ) − 𝜋 (𝑟 2 (𝑟 − ℎ) + (𝑟 − ℎ)3 )
3
3
2

(𝑟 − ℎ)
2
= 𝜋 ( 𝑟 3 + (𝑟 − ℎ) (−𝑟 2 +
))
3
3

2
𝑟 2 2𝑟ℎ ℎ2
= 𝜋 ( 𝑟 3 + (𝑟 − ℎ) (−𝑟 2 + −
+ ))
3
3
3
3
2
2𝑟 2 2𝑟ℎ ℎ2
= 𝜋 ( 𝑟 3 + (𝑟 − ℎ) (−

+ ))
3
3
3
3
2 3 2 3 2 2
1 2 2 2
2 2 ℎ3
= 𝜋 ( 𝑟 − 𝑟 − 𝑟 ℎ + 𝑟ℎ + 𝑟 ℎ + 𝑟ℎ − )
3
3
3
3
3
3
3

𝟏
𝑽 = 𝝅 (𝒓𝒉𝟐 − 𝒉𝟑 )
𝟑

#4. Use the course of cylindrical shells to invent the quantity of the solid
torus indicated in the image adown.

Solution
𝑅+𝑟

𝑉 = ∫ 2𝜋𝑥 . 2√𝑟 2 − (𝑥 − 𝑅)2 𝑑𝑥 ; 𝑢 = 𝑥 − 𝑅
𝑅−𝑟
𝑟

= ∫ 4𝜋(𝑢 + 𝑅)√𝑟 2 − 𝑢2 𝑑𝑢
−𝑟
𝑟

𝑟

= 4𝜋𝑅 ∫ √𝑟 2 − 𝑢2 𝑑𝑢 + 4𝜋 ∫ 𝑢√𝑟 2 − 𝑢2 𝑑𝑢
−𝑟

−𝑟

𝑟

∫ √𝑟 2 − 𝑢2 𝑑𝑢 ; 𝐴𝑝𝑝𝑙𝑦 𝑇𝑟𝑖𝑔 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑜𝑛: 𝑢 = 𝑟𝑠𝑖𝑛(𝑣) ; 𝑑𝑢 = 𝑟𝑐𝑜𝑠(𝑣)𝑑𝑣
−𝑟

2

= ∫ √𝑟 2 − (𝑟𝑠𝑖𝑛(𝑣)) 𝑟𝑐𝑜𝑠(𝑣)𝑑𝑣

= ∫ 𝑟 2 ...


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