Example//Calculate second moment of inertia about y-axis for area enclosed by y=x², y=x+2. Solution // ly = ff x² dA J1=y2 x² = x+2 x2 x 2 = 0 (x-2)(x + 1) = 0 X=2 or x=-1 2 x+2 = [] x²dy dx = [][]*~²dy dx = = SS 2 2 = r² = -1 x+2 -1 x2 (x²y)| dx x² x²(x+2x2)dx = 365 unit of length 2x3 x4 + 4 3 - x5 2 5 Example// Find the centroid enclosed function Y=6x-x². y=x Solution / My Mx X = Y = A A У1=92 6x-x² = x x²-6x+x=0 x2 – 5x = 0 X (x-5)=0 X=0 or x=5 A=ſſ dy dx = √5 √6x¯x² dy dx = √5 (y)| 6x-x² dx = x²x) dx = 125/6 Mx=ff ydA = ff y dy dx 5 1x = √5 (6x- Mx=√ √6x-x² ydy dx = So³ (²) | 6x-x² dx = 5 (6x-x²)2 (x)²) dx = 104.16 2 Also My=625/12 X = My A = Mx Y = = 625 12 125 = 2.5 104.16 A 125/6 = 5 x 2
Example//Calculate second moment of inertia about y-axis for area enclosed by y=x², y=x+2. Solution // ly = ff x² dA J1=y2 x² = x+2 x2 x 2 = 0 (x-2)(x + 1) = 0 X=2 or x=-1 2 x+2 = [] x²dy dx = [][]*~²dy dx = = SS 2 2 = r² = -1 x+2 -1 x2 (x²y)| dx x² x²(x+2x2)dx = 365 unit of length 2x3 x4 + 4 3 - x5 2 5 Example// Find the centroid enclosed function Y=6x-x². y=x Solution / My Mx X = Y = A A У1=92 6x-x² = x x²-6x+x=0 x2 – 5x = 0 X (x-5)=0 X=0 or x=5 A=ſſ dy dx = √5 √6x¯x² dy dx = √5 (y)| 6x-x² dx = x²x) dx = 125/6 Mx=ff ydA = ff y dy dx 5 1x = √5 (6x- Mx=√ √6x-x² ydy dx = So³ (²) | 6x-x² dx = 5 (6x-x²)2 (x)²) dx = 104.16 2 Also My=625/12 X = My A = Mx Y = = 625 12 125 = 2.5 104.16 A 125/6 = 5 x 2
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Please (re)solve the questions below in some detail
![Example//Calculate second moment of inertia about y-axis
for area enclosed by y=x², y=x+2.
Solution //
ly
= ff x² dA
J1=y2
x² = x+2
x2 x 2 = 0
(x-2)(x + 1) = 0
X=2 or x=-1
2 x+2
= [] x²dy dx = [][]*~²dy dx =
=
SS
2
2
= r² =
-1
x+2
-1 x2
(x²y)| dx
x²
x²(x+2x2)dx
= 365 unit of length
2x3
x4
+
4
3
-
x5 2
5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b509b-2aec-4740-924b-bdebd5fee3b4%2F0a09331a-3f5c-4638-8c14-f056de97f78f%2Fmtyzt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example//Calculate second moment of inertia about y-axis
for area enclosed by y=x², y=x+2.
Solution //
ly
= ff x² dA
J1=y2
x² = x+2
x2 x 2 = 0
(x-2)(x + 1) = 0
X=2 or x=-1
2 x+2
= [] x²dy dx = [][]*~²dy dx =
=
SS
2
2
= r² =
-1
x+2
-1 x2
(x²y)| dx
x²
x²(x+2x2)dx
= 365 unit of length
2x3
x4
+
4
3
-
x5 2
5
Transcribed Image Text:Example// Find the centroid enclosed function
Y=6x-x². y=x
Solution /
My
Mx
X
=
Y =
A
A
У1=92
6x-x² = x
x²-6x+x=0
x2 – 5x = 0
X (x-5)=0
X=0 or x=5
A=ſſ dy dx = √5 √6x¯x² dy dx = √5 (y)| 6x-x² dx =
x²x) dx = 125/6
Mx=ff ydA = ff y dy dx
5
1x = √5 (6x-
Mx=√ √6x-x² ydy dx =
So³ (²) | 6x-x² dx =
5 (6x-x²)2
(x)²) dx = 104.16
2
Also My=625/12
X =
My
A
=
Mx
Y =
=
625
12
125
= 2.5
104.16
A 125/6
= 5
x
2
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