Use a triple integral to önd the volume of the solid enclosed by the cylinder x 2 + z 2 = 4 and the planes y = 1 and y + z = 4.

the volume is V=12π

Step-by-step explanation:

using cylindrical coordinates

x= rsin θ

z= rcos θ

y=y

therefore

y+z=4 → y= 4-z = 4-r cos θ

also from x²+z²=4 →  -2≤x≤2 , -2≤z≤2

therefore since y= 4-z  → 6≤y≤2 → it does not overlap with the plane y=1

V=∫∫∫dV = ∫∫∫dxdydz = ∫∫∫rdθdrdy = ∫∫rdθdr   [(y=4-r cos θ,y=1) ∫ dy] =

∫∫[(4-rcosθ) - 1]rdθdr =  ∫∫(3-rcosθ) rdθdr = ∫dθ [r=2,r=0] ∫(3r-r²cosθ) dr

∫ (3/2* 2²- 2³/3 cosθ) dθ =[θ=2π, θ=0] ∫ (6-8/3 cosθ) dθ = 2π*6 - 8/3 sin0 = 12π

thus

V= 12π

to verify it, the volume should not be bigger than the volume if the cross section was a square and thus the volume enclosed would be:

V = [(2-(-2)]² * (6-2) /2 + [(2-(-2)]²  * (2-1) = 4³/2 + 4²*2 = 64 > 12π

stops on an even number.

answer: the answer is ann