find the volume of the solid obtained by rotating the region y=8x^3 , y=0 , x=1 ; about x=2

Answer 1

Use the shell method.

Take a vertical cross section of the region. The width of the cross section is dx, and the height is 8x³.

Rotate the cross section about the line x = 2. The height of the shell is 8x³, the radius is 2 - x, and the thickness is dx. .

Volume of shell = 2π(2-x)(8x³)dx = 2π(16x³ - 8x⁴)dx

Volume of solid = 2π∫_{(from 0 to 1)}[16x³ - 8x⁴]dx = 2π(4x⁴ - (8/5)x⁵)_{(from 0 to 1)} = 24π/5

Answer 2

Bounded by y = 8x³, y = 0, x = 1& revolved about the line x = 2

The best thing to do is to draw the figure out. If you draw your rectangular strip vertically, you'll see that you are going to have a cylinder (or shell). So, you'll be using the formula

V = 2π∫_a^b r(h)dx

In this case, the radius r, is the distance from your line of rotation to the vertical strip (which is at a random x value). Thus, the radius r = 2 - x

The height of the strip is the height of the y - value of the function, thus h = 8x³

Hence we have,

V = 2π∫₀¹(2 - x)(8x³) dx

= 2π*8∫₀¹ (2x³ - x⁴) dx <--------The 8 is a constant, so I just pulled it out.

= 16π(1/2 x⁴ - 1/5 x⁵) |₀¹

= 16π [(1/2 - 1/5) - 0]

= 16π (3/10)

= 24π/5

Hope this helps

Mr. K