Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=0 , y=cos (7x) , x=π/14 , x=0 about the axis x=3

π/49·(44 -π)

Step-by-step explanation:

Using the "shell" method, a differential of volume is the product of the area of a cylindrical shell and its circumference around the axis of rotation. Here, that area is ...

dA = y·dx = cos(7x)dx

The radius will be the difference between x and the axis of rotation, x=3, so is ...

r = 3 -x

Then the differential of volume is ...

dV = 2πr·dA = 2π(3-x)cos(7x)dx

The volume will be the integral of this over the limits x ∈ [0, π/14].

∫dV = 6π·∫cos(7x)dx -2π·∫x·cos(7x)dx . . . from 0 to π/14

= (6/7)π·sin(7x) -(2/49)π·(cos(7x) +7x·sin(7x)) . . . from 0 to π/14

= (6/7)π(1 -0) -(2/49)π((0 -1) +(π/2-0))

= π(42/49 +2/49 -π/49)

= (π/49)(44 -π) . . . . cubic units . . . . . approx 2.61960 cubic units

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)