Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2006-02-12 21:56:04

johny
Member

Offline

Integration problem!!!!!

Hello guys, well recently i have been really struggling with integration. Heres the question:

A piece of machinery has a cutter which is cooled by a water soluble oil. The cooling solution is held in a reservoir in the shape of frustrum of a cone.

We need to find the volume by integration of the reservoir.

Heres the pic:

johny
Member

Offline

Pic:

#3 2006-02-13 01:44:27

irspow
Power Member

Offline

Re: Integration problem!!!!!

I prefer to use the method of cylindrical shells for a problem like this.  This method can be thought of as the summation circumferences multiplied by the height.  It is like spinning an area around the axis of choice. It looks like this;

∫2πxf(x)dx...note that shell method spins around the opposite axis as the variable.  The disk or washer methods spins around the same axis as the variable so you would have to use f(y) functions for those.

First, we need f(x) for the boundary or side of the object.

Note the end points of the right side of the frustrum.  (3,30),(8,0)

If you are taking calculus, I suppose that you know how to make a line equation from this.

For the sake of clarity anyway;

(y1 - 30) = (dy/dx)(x1 - 3),   dy/dx from the points above gives -30/5 = -6

(y1 - 30) = -6x1 + 18,   we will call this ys (side)

So, ys = 48 - 6x,  but notice that we are only going to use this function when x goes from 3 to 8

Before x passes 3 we simply have a cylinder with a radius of 3 and height of 30.

Here y = 30 for x from 0 to 3, we will call this yc (center)

So using the shell method from way above for both functions with their proper limits of integration gives us;

∫2πx(30)dx] from 0 to 3  + ∫2πx(48 - 6x)dx] from 3 to 8 =

2π {∫30x dx]0,3 + ∫48x - 6x² dx]3,8}

2π {(15x²)]0,3 + (24x² - 2x³)]3,8}

2π [(135) + (1536 - 1024 - 216 + 54)]

2π (485) =

V = 970π cm³

edit*

I had to correct a silly error concerning multiplication!

Last edited by irspow (2006-02-15 10:59:15)

#4 2006-02-14 04:50:32

johny
Member

Offline

Re: Integration problem!!!!!

Thanks once again irspow!!!!

#5 2006-02-14 10:34:04

irspow
Power Member

Offline

Re: Integration problem!!!!!

Hey johny, did you ever find out what the deal with the wipers was?  It has bothered me slightly that everything was not quite right with the final solution in that thread.  Was the area posted correct?  If so, what was wrong with the stipulated conditions?  If not, what did everyone miss?

If you could shed any light on that, it would be greatly appreciated by me.  Thanks.

#6 2006-02-14 19:36:43

johny
Member

Offline

Re: Integration problem!!!!!

Well i havent checked  anything yet, as soon as i find out i will post it in the same topic