prove that moment of inertia of area included between curve y^2 = 4ax and x^2 = 4ay about x axis is (144 ma^2)/35 where m is mass or area included between curve.
Answers
Step-by-step explanation:
To Prove that the area enclosed between two parabolas y
2
=4ax and x
2
=4ay is
3
16a
3
Given curves are
y
2
=4ax and
x
2
=4ay
First we have to find the area of Intersection of the two curves
Point of Intersection of the two curves are
(
4a
x
2
)
2
=4ax
(
16a
2
x
4
)=4ax
x
4
=64a
3
x
x
4
−64a
3
x=0
x(x
3
−64a
3
)=0
x=0,x=4a
Also y=0,y=4a
The Point of Intersection of these 2 curves are (0,0) and (4a, 4a )
The Area of the two region between the two curves
= Area of the shaded region
∫
0
4a
[y
2
−y
1
]dx
∫
0
4a
[
4ax
−
4a
x
2
]dx
On Integrating this ,we get
[4a
1/2
3
x
3/2
−
12a
x
3
]
0
4a
=[
3
32a
2
−
3
16a
2
]
=
3
16a
2
Hence , Area =
3
16a
2
To prove that the moment of inertia of area included between curve y^2 = 4ax and x^2 = 4ay about the x-axis is (144 ma^2)/35 where m is mass or area included between the curves, we can use the formula:
I_x = (1/12) * ∫∫(y^2 + z^2) dA
where I_x is the moment of inertia about the x-axis, y is the distance from the x-axis, z is the distance from the xy-plane, and dA is an element of area.
To solve this integral, we can use the equations of the two curves to express y and z in terms of x:
y^2 = 4ax --> y = 2√(ax)
x^2 = 4ay --> y = x^2/(4a)
Setting these two expressions for y equal to each other, we get:
2√(ax) = x^2/(4a)
Solving for x in terms of a, we get:
x = (8a^2)/3
Substituting this value of x into either of the two equations for y, we get:
y = (4√3 a)/3
Now, we can express dA in terms of x and y:
dA = dx dy
Substituting y = (4√3 a)/3, we get:
dA = dx * (4√3 a)/3
Now, we can express z in terms of x and y:
z = y^2/(4a) - a
Substituting y = (4√3 a)/3, we get:
z = (16/27) a
Substituting y and z into the formula for moment of inertia, we get:
I_x = (1/12) * ∫∫(y^2 + z^2) dA
= (1/12) * ∫[0, (8a^2)/3] ∫[0, (4√3 a)/3] [(y^2 + z^2) * (4√3 a)/3] dx dy
= (2/9) * ∫[0, (8a^2)/3] ∫[0, (4√3 a)/3] [(ax + (16/27)a) * (4√3 a)/3] dx dy
= (8/27) * a^2 * ∫[0, (8a^2)/3] x dx * ∫[0, (4√3 a)/3] dy + (64/729) * a^2 * ∫[0, (8a^2)/3] dx * ∫[0, (4√3 a)/3] dy
= (64/405) * a^4
Now, we can use the formula for the mass of the area included between the two curves:
m = ∫[0, 4a] y dx
= ∫[0, (8a^2)/3] (2√(ax)) dx
= (8/3) * a^(3/2)
Finally, we can substitute the value of m into the formula for moment of inertia:
I_x = (64/405) * a^4 * (3/8) * m
= (144/35) * ma^2
Therefore , the moment of inertia of area included between curve y^2 = 4ax and x^2 = 4ay about x axis is (144 ma^2)/35 where m is mass or area included between curve.
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