Ind the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 4 − 3 2 x, y = 0, x = 0, x = 1; about the x-axis

V=258 πvu

Step-by-step explanation:

Knowing that to find the solid of revolution is V=πR²h, where R² = y = f(x) and  h = a differential of x, we have that (according to the graph)

dV=π(4-32x)²dx, integrating on both sides    ∫dV=π∫(4-32x)²dx;  

let's solve ∫(4-32x)²dx = ∫(16-256x+1024x²)dx = 16∫dx-256∫xdx+1024∫x²dx =

16x-128x²+(1024/3)x³ evaluated between 0≤x≤1/8 plus 1/8∠x≤1, also

V=π/8(16.-128.+(1024/3))+π(16.1-128.1+(1024/3).1 = π/8(48-384+1024)/3+π(16.1-128.1+(1024/3).1 = π688/24+ π688/3 =

258 πvu

Note:

vu = volume units

Due to "inconvenience" to visualize graph 1 in its real scale its scale was modified in graph 2 and 3 respectively