Calculus Help! - Revolving an area about a line! 10pts!!

Find the volume of the solid for the region bounded by the lines y=6-x. y=0, x=0 revolved about the line x=6.

Can anyone also verify the shape, is it a trapezoid from 0 to 6 and a symmetrical trapezoid from 6 to 12. Or is a cone/triangle like shape that's centered at 6?

Thanks.

It's the triangle cone thing centered at 6. The axis of rotation is vertical, so the shell method requires slicing vertically, implying dx. Note that the lines intersect at 0 and 6.

V = 2π ∫ rh(x) dx from 0 to 6

First determine the radius, r. When x = 0 the radius is 6 and when x = 6 the radius is 0, so the radius is 6 - x.

Now determine the height h(x). The height is defined by the line y = 6 - x, so the height is just 6 - x. Now sub all of that in:

V = 2π ∫ (6 - x)(6 - x) dx from 0 to 6
V = 2π ∫ (6 - x)^2 dx from 0 to 6

let u = 6 - x
du = - dx

V = - 2π ∫ (u)^2 du from 0 to 6
V = - 2π[(1/3)u^3] from 0 to 6
V = (-2/3)π[(6 - x)^3] from 0 to 6
V = - (-2/3)π[(6 - 0)^3] <-- Note that the upper limit of 6 will be 0
V = 144π

Done!

It is a washer.
Volume = int(3.1416*R^2) - (3.1416* r^2).dx| from 0 to 6

R =outer radius
r = inner radius