Can you have an 'All Real Numbers' solution for inequalities? If so, give an example.
1 answer:
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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
