Question

A solid lies between planes perpendicular to the​ x-axis at x = -2 and x = 2. The​ cross-sections perpendicular to the​ x-axis between 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.

Answer 1

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