Calculus Help! - Revolving an area about a line! Need help showing the work! 10 pts!!!!

Find the volume of the solid for the region bounded by the lines y=6-x. y=0, y=4, x=0 revolved about the line x=6.

Can anyone also verify the shape, is it a rectangle with a interior triangle that intercept at x=6. Or is a triangle/cone that have it's center at x=0.

Thanks.

Answer is 368/3pi, can someone show me how to get that?

Third time's the charm eh ;)

The axis of rotation is vertical, but the washer method will be easier for this, so slice horizontally, implying dy. Now the limits are from 0 to 4.

V = ∫ A(y) dy from 0 to 4

A(y) = πR^2 - πr^2
A(y) = π(R^2 - r^2)

So first determine the outer radius, R. The outer radius is defined by the line x = 0, and since x = 0 is always 6 away from x = 6, R = 6.

Now determine the inner radius, r. The inner radius is defined by the line x = 6 - y (rearranging the equation for x.) When y = 0 the radius is 0 and when y = 4 the radius is 4, so r = y. Note that we didn't actually need the equation, but it's useful to have it just in case. Now sub those in:

A(y) = π(R^2 - r^2) = π((6)^2 - (y)^2) = π(36 - y^2)

V = π ∫ (36 - y^2) dy from 0 to 4
V = π[36y - (1/3)y^3] from 0 to 4
V = π[36(4) - (1/3)(4)^3] <-- Note that the lower limit of 0 will be zero
V = (368/3)π <--- yay!

Done!

P.S. Don't worry about that typo. It happens to the best of us...

P.S.S. If you want a diagram of the lines:

http://www.wolframalpha.com/input/?i=plo…

Edit:

The area that we're integrating is the one that looks like this:

_______
|............\
|.............\
|..............\
|...............\
|................\
|.................\
|..................\

It's the triangle, but with the very top cut off.

There's only one area to integrate - the area that I drew is IT. There isn't anything else. You revolve that area around x = 6 which, yes, would mean there's another cross section from 6 to 12, but you don't need to worry about that. The integration takes care of it.

Report Abuse