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.

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

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!