A solid lies between planes perpendicular to the x-axis at x = -2 and x = 2. The cross-sections perpendicular to the x-axis b
etween these planes are squares whose bases run from the semicircle y = -√(4 - x²) to the semicircle y = √(4 - x²).
Find the volume of the solid.
Find the volume of the solid.
1 answer:
Answer:
V = 128π/3 vu
Step-by-step explanation:
we have that: f(x)₁ = √(4 - x²); f(x)₂ = -√(4 - x²)
knowing that the volume of a solid is V=πR²h, where R² (f(x)₁-f(x)₂) and h=dx, then
dV=π(√(4 - x²)+√(4 - x²))²dx; =π(2√(4 - x²))²dx ⇒
dV= 4π(4-x²)dx , Integrating in both sides
∫dv=4π∫(4-x²)dx , we take ∫(4-x²)dx and we solve
4∫dx-∫x²dx = 4x-(x³/3) evaluated -2≤x≤2 or too 2 (0≤x≤2) , also
∫dv=8π∫(4-x²)dx evaluated 0≤x≤2
V=8π(4x-(x³/3)) = 8π(4.2-(2³/3)) = 8π(8-(8/3)) =(8π/3)(24-8) ⇒
V = 128π/3 vu
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Step-by-step explanation: For two events A and B, we are given the following probabilities :
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Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.