The correct answer for the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y = 4x is 128/3 * π.
Volume of Solid of Revolution by Shell method is given by
V = 2π * integrate x(height) dx Here, height = 4x-x2
(1)& x-varies from x = 0 to x = 4 then from eqn(1) V = 2π * integrate x(4x - x ^ 2) dx from x = 0 to 4 = 2π * integrate (4x ^ 2 - x ^ 3) dx from x = 0 to 4
Basic Rule(1) ∫ x^n dx =x^ n+1/ n+1
V=2 π [4((x ^ 3)/3) - (x ^ 4)/4] 0 ^ 4 =2 π[ 4/3 x^ 3 - x^ 4/4 ] 0 ^ 4
V = 2π [4/3 * 4 ^ 3 - (4 ^ 4)/4} - 0]
V = 128/3 * π.
Learn more about area under curve here :-
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Neither the speed nor the distance of a falling object is linearly related to time.
When smoke particles pass between the source of radiation and the detector the drop in radiation will be sufficient to tigger the alarm. <span />
According to law of conservation of energy,
<span>Energy can neither be constructed nor be destroyed but can be transformed from one form to another.
</span>
<span>At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
</span><span>Therefore total energy at point b = potential energy = 711 J.... i
</span>
<span>At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
</span>Therefore total energy at point a = kinetic energy ---- ii
<span>From i and ii,
</span>Kinetic energy = potential energy = 711 J.(Conserving energy)
Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!
Answer:
Explanation:
This is case of interference in thin films
for constructive interference in thin film the condition is
2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .
2 x 1.28 x t = λ/2 , if n = 0
2 x 1.28 x t = 605 /2
t = 118.16 nm .
the minimum non-zero thickness of the oil film required = 118.16 nm.