Find the volume generated by revolving about the x-axis y = 6x-x 2 , y = 0?

Answer 1

what answerer 1 said...

Answer 2

if you are revolving around the x axis then you can slice your discs vertically to get a thin width of dx.

first find the volume of one such thin disk then integrate along the x axis.

volume of disc = Area of circle * dx
= pi*r^2 * dx

since the radius of your disk is set by y values so

volume of disc = pi*(6x-x^2)^2 *dx

please expand that expression.

then volume of the entire solid is

integ(pi*(6x-x^2)^2 *dx)

if you have bounds, you may first integrate then substitute the numbers in.

*note i don't like to give away answers so i hope you don't simply copy down the first reply but work through it yourself

Answer 3

y = 0 doesn't really add anything (this is the x-axis).

y = 6x - x^2 = 0 when x = 0, 6.

So the volume is
V = π∫y^2 dx
= π∫(0 to 6) (6x - x^2)^2 dx
= π∫(0 to 6) (36x^2 - 12x^3 + x^4) dx
= π [12x^3 - 3x^4 + x^5 / 5] (0 to 6)
= π (12 . 6^3 - 3 . 6^4 + 6^5 / 5 - 0)
= π (2592 - 3888 + 7776/5)
= 1296π/5
= 814.3 (1 d.p.).