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Work done on a frustum

Pindrought

New member
Mar 3, 2014
15
Okay so please bear with me. I'm dying here.

Here's a diagram of the situation I have.




I set up the graph to be like so for what I would be rotating to get the volume. (Edit: I meant to put y = 2+1/5x)


I'm really stuck here though. I know I need to pump the water to 11ft from it's current location so I could get the distance of the water from the top with (11-x) for where i'm integrating. I've searched up several of these types of problems and i'm having difficulty getting the work concept through my head of how to approach these.

Thanks so much for reading.(Tongueout)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: Work done on a frustrum

Let's start by finding $c(x)$: the ratio of the linear size (e.g., the diameter if the container is round) of the section at height $x$ to the linear size of the bottom. This ratio is $1$ when $x=0$, and it is $4/2=2$ when $x=10$. It also changes linearly. Therefore, $c(x)=1+x/10$. Now, the ratio of the areas of similar figures is the square of the ratio of their linear sizes. Therefore, the ratio of the section's area at height $x$ to the area of the bottom is $c^2(x)$. Suppose the area of the bottom is $S$; then the area at height $x$ is $Sc^2(x)$.

Now find the volume of the layer of water of depth $dx$ at height $x$. Multiply it by water's density to find its mass. Multiply it by $g$ to find its weight. Multiply it by the distance it has to travel to find work. Finally, integrate this expression from $x=0$ to $x=5$.
 

shamieh

Active member
Sep 13, 2013
539
Re: Work done on a frustrum

I saw this problem in the book as well. Is this what Pindrought's integral should look like? \(\displaystyle \int^5_0 \pi * \frac{1}{5}((10 - c))^2 dx * 2000g\) ?
 

Pindrought

New member
Mar 3, 2014
15
Re: Work done on a frustrum

\(\displaystyle \int^5_0 \pi (2 + \frac{1}{5} x)^2 * (11 - x) * 2000 \, dx\)

I was thinking i'd set it up like this. Would someone mind just telling me if this is wrong so I can try again if it is?
 

shamieh

Active member
Sep 13, 2013
539
Re: Work done on a frustrum

Actually, now I'm getting this \(\displaystyle \int^{10}_0 \pi(\frac{1}{2}((10 - x))^2 - 2000g \, dx\)
 

shamieh

Active member
Sep 13, 2013
539
Re: Work done on a frustrum

anyone know the answer to this?
 

Pindrought

New member
Mar 3, 2014
15
Re: Work done on a frustrum

Let's start by finding $c(x)$: the ratio of the linear size (e.g., the diameter if the container is round) of the section at height $x$ to the linear size of the bottom. This ratio is $1$ when $x=0$, and it is $4/2=2$ when $x=10$. It also changes linearly. Therefore, $c(x)=1+x/10$. Now, the ratio of the areas of similar figures is the square of the ratio of their linear sizes. Therefore, the ratio of the section's area at height $x$ to the area of the bottom is $c^2(x)$. Suppose the area of the bottom is $S$; then the area at height $x$ is $Sc^2(x)$.

Now find the volume of the layer of water of depth $dx$ at height $x$. Multiply it by water's density to find its mass. Multiply it by $g$ to find its weight. Multiply it by the distance it has to travel to find work. Finally, integrate this expression from $x=0$ to $x=5$.
Am I misunderstanding or did you interpret it as my top radius was 2 and my bottom radius was 1?

It's supposed to be that my upper radius is 4 and my bottom radius is 2.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Work done on a frustrum

I would first look at slicing the contents of the tanks into thin disks, that is to slice perpendicularly to the axis of symmetry, where each point in the disk has the same gravitational potential energy. Now, we want to know the radius of an arbitrary disk, as a function of its distance from the bottom of the tank. We know that as we move away from the bottom of the tank, the radius of the tank increases linearly.

So, if we orient a $y$-axis through the axis of symmetry of the tank, and put the origin of this axis at the bottom of the tank, we know two points on this linear radius function:

\(\displaystyle (y,r)=(0,2),\,(10,4)\)

Now, knowing two points on a line, how do we go about finding the equation of the line?
 

Pindrought

New member
Mar 3, 2014
15
Re: Work done on a frustrum

I would first look at slicing the contents of the tanks into thin disks, that is to slice perpendicularly to the axis of symmetry, where each point in the disk has the same gravitational potential energy. Now, we want to know the radius of an arbitrary disk, as a function of its distance from the bottom of the tank. We know that as we move away from the bottom of the tank, the radius of the tank increases linearly.

So, if we orient a $y$-axis through the axis of symmetry of the tank, and put the origin of this axis at the bottom of the tank, we know two points on this linear radius function:

\(\displaystyle (y,r)=(0,2),\,(10,4)\)

Now, knowing two points on a line, how do we go about finding the equation of the line?
That's what I thought I did when I said y=2+(1/5)x which I plugged in for the radius. Still can't figure out where i'm messing up.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: Work done on a frustrum

Someone has to say it, but it's frustum, not frustrum (Tongueout) (pedantic, perhaps, but will possibly save you some embarrassment later on)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Work done on a frustrum

Okay good, I am trying to build up one piece at a time, so we know exactly what we're doing before jumping to the integral. :D

So, we have:

\(\displaystyle r=\frac{1}{5}y+2\)

So, we then move to the volume of the arbitrary disk. It is given by:

\(\displaystyle dV=\pi r^2\,dy=\pi\left(\frac{1}{5}y+2 \right)\,dy=\frac{\pi}{25}(y+10)^2\,dy\)

Now, we want to find the weight of the slice. Let's let $\rho$ be the weight density of the fluid per unit volume. We know that density, weight $w$ and volume $V$ are related by:

\(\displaystyle \rho=\frac{w}{V}\therefore\,w=\rho V\)

And so the weight of the slice is:

\(\displaystyle w=\rho\,dV=\frac{\rho\pi}{25}(y+10)^2\,dy\)

Now, you want to use the face that the work required to raise the slice the required height is the product of the force (weight) and the distance $d$ over which it is lifted, which is:

\(\displaystyle d=11-y\)

And so the work required to lift the slice is:

\(\displaystyle dW=w\cdot d=\frac{\rho\pi}{25}(11-y)(y+10)^2\,dy\)

Now it is time to add up all the increments of work to get the total:

\(\displaystyle W=\frac{\rho\pi}{25}\int_0^5 (11-y)(y+10)^2\,dy\)

So now, you want to expand the integrand and then apply the FTOC. What do you find?
 

Pindrought

New member
Mar 3, 2014
15
Re: Work done on a frustrum

Okay good, I am trying to build up one piece at a time, so we know exactly what we're doing before jumping to the integral. :D

So, we have:

\(\displaystyle r=\frac{1}{5}y+2\)

So, we then move to the volume of the arbitrary disk. It is given by:

\(\displaystyle dV=\pi r^2\,dy=\pi\left(\frac{1}{5}y+2 \right)\,dy=\frac{\pi}{25}(y+10)^2\,dy\)

Now, we want to find the weight of the slice. Let's let $\rho$ be the weight density of the fluid per unit volume. We know that density, weight $w$ and volume $V$ are related by:

\(\displaystyle \rho=\frac{w}{V}\therefore\,w=\rho V\)

And so the weight of the slice is:

\(\displaystyle w=\rho\,dV=\frac{\rho\pi}{25}(y+10)^2\,dy\)

Now, you want to use the face that the work required to raise the slice the required height is the product of the force (weight) and the distance $d$ over which it is lifted, which is:

\(\displaystyle d=11-y\)

And so the work required to lift the slice is:

\(\displaystyle dW=w\cdot d=\frac{\rho\pi}{25}(11-y)(y+10)^2\,dy\)

Now it is time to add up all the increments of work to get the total:

\(\displaystyle W=\frac{\rho\pi}{25}\int_0^5 (11-y)(y+10)^2\,dy\)

So now, you want to expand the integrand and then apply the FTOC. What do you find?
Thanks for clearing that up. I only have a question about one thing that confuses me.

for density, \(\displaystyle \rho\), should it be just the density of water (62.5lb/ft^3) or should it be the density of water*gravity (62.5lb/ft^3)*32 ?

I would think it is just 62.5lb/ft^3, but before for some reason I had thought that it would be gravity*density which would have been 62.5*32 which is how I was getting the *2000

Thanks a lot for the help
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are correct, the density:

\(\displaystyle \rho=62.5\frac{\text{lb}}{\text{ft}^3}\)

is a weight density, since it is weight (lb) per unit volume (cu. ft.). :D