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Given The Equation 2x-Y 12 and 4x Y: A) - Statement of The Problem

Given the equations 2x-y=12 and 4x=y^2, find the area and volume bounded by the curves. To find the area: 1) Find the points of intersection which are (9,6) and (4,-4) 2) Integrate to find the area between the curves, which is 125 square units To find the volume: 1) Set up integrals to find the volume of revolving the area about the y-axis 2) Integrate and simplify to find the volume is 1000π cubic units

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Kesia Almodal
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64 views5 pages

Given The Equation 2x-Y 12 and 4x Y: A) - Statement of The Problem

Given the equations 2x-y=12 and 4x=y^2, find the area and volume bounded by the curves. To find the area: 1) Find the points of intersection which are (9,6) and (4,-4) 2) Integrate to find the area between the curves, which is 125 square units To find the volume: 1) Set up integrals to find the volume of revolving the area about the y-axis 2) Integrate and simplify to find the volume is 1000π cubic units

Uploaded by

Kesia Almodal
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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AREA

a). Statement of the problem

Given the equation 2x-y=12 and 4x=y², find the area bounded by the two curves

b) Graphical Representation

c) Solution (Computation)

Point of intersection:
x=x The line ; 2x-y=12
12 𝑦 𝑦2
+ = y=2x-12
2 2 4
𝑦 𝑦2
6 + = Y-INTERCEPT : let x= 0
2 4
𝑦2 𝑦
+ –6=0 y= 2(0)-12
4 2
𝑦 2 −2𝑦−24
= 0 y= 12
4
𝑦 2 − 2y − 24 = 0 X-INTERCEPT : let y=0
(𝑦 − 6) (y + 4)=0 y= 2x-12
y=6 y=−4 0= 2x-12
x= 6
when ; y=6 when; y = −4
12+6 12 + (−4)
𝑥= 𝑥=
2
2
𝒙=𝟗
𝒙=𝟒
(9 , 6)
(4 , -4)

6
12 + 𝑦 𝑦²
𝐴=∫ ( − ) 𝑑𝑦
−4 2 4

1 6 1 6
𝐴= ∫ (12 + 𝑦)𝑑𝑦 − ∫ (𝑦²)dy
2 −4 4 −4

1 y² 1 𝑦3
6
𝐴 = 2 (12𝑦 + 2 ) − 4 ( 3 )]−4

𝑦² 𝑦3 6
𝐴 = 6𝑦 + − 12]−4
4

(6)² (6)3 (−4)2 (−4)3


𝐴 = 6(6) + − − [6(−4) + − ]
4 12 4 12

216 64
𝐴 = 36 + 9 − − [−24 + 4 + ]
12 12
216 64
𝐴 = 45 − − [−20 + ]
12 12
216 − 64
𝐴 = 45 + 20 −
12
280
𝐴 = 65 −
12
70
𝐴 = 65 −
3
195 − 70
𝐴=
3
125 2
𝐴= 𝑢
3
d) Proving

= 10𝑥9 6 𝑦2
‫׬‬0 𝑑𝑦
4
𝑥90 𝑢2
1 6 2
1 = ∫ 𝑦 𝑑𝑦
4 0
𝐴△ = 𝑏ℎ
2
1 𝑦3 6
1 = ( )]
𝐴△ = (10𝑥5) 4 3 0
2
1 𝑦3 6
𝐴△ = (50) = ]
2 12 0

𝐴△ = 25 𝑢2 ∎ (6)3
= −0
12
= 18 𝑢2 ∎

0
𝑦2 64
∫ 𝑑𝑦 [90 − ൬25 + 1 + ൰]
−4 4 12

𝑦3 0
= ]
12 −4 500 125 2
= 𝑜𝑟 𝑢 ∎
12 3
(−4)3 64 2
=0−[ ]= 𝑢 ∎
12 12
VOLUME
a). Statement of the problem

Given the equation 2x-y=12 and 4x=y², find the volume generated by revolving
the area bounded by the given curves about y-axis

b) Graphical Representation

c) Solution (Computation)
Given curves: 2x-y=12 and y²= 4x

2
6
12 + 𝑦 2 𝑦2
𝑉 = ∫ 𝜋 (൬ ൰ − ( ) ) 𝑑𝑦
4 2 4

2
6
12 + 𝑦 2 𝑦2
𝑉 = 𝜋 [∫ (൬ ൰ − ( ) ) 𝑑𝑦]
4 2 4

2
6
12 + 𝑦 2 6
𝑦2
𝑉 = 𝜋 [∫ ൬ ൰ 𝑑𝑦 − ∫ ( ) 𝑑𝑦]
−4 2 −4 4

1 6 1 6
𝑉 = 𝜋 [ ∫ 144 + 24𝑦 + 𝑦 2 𝑑𝑦 − ∫ 𝑦 4 𝑑𝑦]
4 −4 16 −4
1 𝑦3 1 𝑦5 6
𝑉 = 𝜋 [ (144𝑦 + 12𝑦 2 + 𝑑𝑦 − ( ))]
4 3 16 5 −4

𝑦3 𝑦5 6
𝑉 = 𝜋 [36𝑦 + 3𝑦 2 + − ]]
12 80 −4

(6)3 (6)5 (−4)3 (−4)5


𝑉 = 𝜋 [36(6) + 3(6)2 + − − (36(−4) + 3(−4)2 + − )]
12 80 12 80

216 7776 64 1024


𝑉 = 𝜋 [216 + 108 + − − ൬−144 + 48 − + ൰]
12 80 12 80

216 7776 64 1024


𝑉 = 𝜋 [(324 + − − ൬−144 + 48 − + ൰)]
12 80 12 80

216 7776 64 1024


𝑉 = 𝜋 [(324 + − − ൬−96 − + ൰)]
12 80 12 80

216 + 64 7776 − 1024


𝑉 = 𝜋 [420 + − ]
12 80
70
𝑉 = 𝜋 [420 + − 110]
3
1260 + 70 − 330
𝑉 = 𝜋[ ]
3
1000𝜋
𝑉= 𝑢𝑛𝑖𝑡𝑠 3
3

Almodal Kesia E. Mr. Dela Cruz

11-STEM 4 Basic Calculus

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