Volume of a Frustum - Get Help Now

[archaic]

EDIT: I thought I was in the math section for homework, sorry!
My work is wrong, I don't see why though. Help much appreciated :)

 

[BvU]

What's the problem ?

 

[archaic]

What's the problem ?

$R^2$ would yield $r^2_2-2r_2r_1+r^2_1$ but it should be $r^2_2+r_2r_1+r^2_1$.

 

[BvU]

A problem statement tells us what is given and what is asked for ...

 

[Mark44]

A problem statement tells us what is given and what is asked for ...

Yes. @archaic, please provide a description of the problem you're trying to solve.

 

[archaic]

A problem statement tells us what is given and what is asked for ...

A problem statement tells us what is given and what is asked for ...

Calculate the volume of a frustum.

 

[Mark44]

Calculate the volume of a frustum.

Just any old frustum? What are the dimensions of the frustum whose volume you want to find?
Your drawing isn't much help to your cause. A better drawing, showing the x and y axes, would be helpful. It would also be helpful to draw a sketch of the cone frustum showing the typical volume element.

I've done the problem, and am getting the right answer. I used horizontal disks. Your work appears to use cylindrical shells. It's possible you're not getting the right answer because there's a mismatch between your equation for the sloping edge and what you're using as your integrand. This is where it's important to have a drawing that identifies points with coordinates.

 

[archaic]

Just any old frustum?

Yes.

 

[archaic]

Your work appears to use cylindrical shells.

I have actually done it using the shell method by considering $z=f(r)$ and the result was correct.

 

[Mark44]

So do you still have a question? In both images you posted, you show $V = \frac{\pi R^2h}{3}$. This is the volume of a cone of radius R and height h, but it's not the volume of a frustum of a cone.

 

[archaic]

So do you still have a question? In both images you posted, you show $V = \frac{\pi R^2h}{3}$. This is the volume of a cone of radius R and height h, but it's not the volume of a frustum of a cone.

Yes! I don't understand why that integral gives me that result while the setup is one for a frustum? Bear in mind that $R=r_2-r_1$.

 

[Mark44]

Yes! I don't understand why that integral gives me that result while the setup is one for a frustum? Bear in mind that $R=r_2-r_1$.

It's because your equation for r is wrong. You have $r = \frac R h z = \frac {r_2 - r_1} h z$, which is wrong.
You might have confused yourself by using r and z as coordinate axes.

Based on your drawing, which isn't the easiest to read, the coordinates for the two endpoints of the sloping segment are $(H - h, r_1)$ and $(H, r)$.
Your value for the slope of this line is correct, but your equation for r (really y) isn't. Use the point-slope form of the equation of a line to get the correct equation.

 

[Mark44]

You've made two separate attempts on this problem. In the second try, you have the equation of the line that is the slant edge of the frustum wrong, and I believe that's why your first attempt also produced an incorrect result.

 

[archaic]

No, I have $r=\frac{R}{h}z$

 

[Mark44]

No, I have r=Rhzr=Rhz

Right. I omitted z in my post. What I meant to write is "You have $r = \frac R h z = \frac {r_2 - r_1} h z$, which is wrong."

 

[archaic]

Right. I omitted z in my post (now edited).

How is that rate of change wrong though? You have $\Delta r=R$ for $\Delta z=h$.

 

[Mark44]

How is that rate of change wrong though? You have $\Delta r=R$ for $\Delta z=h$.

As I already said in post #12, your slope for the slant line is correct, but your equation for the line is incorrect.

 

[archaic]

As I already said in post #12, your slope for the slant line is correct, but your equation for the line is incorrect.

I have $r=mz+b$ and $r(0)=0$, so $r=\frac{R}{h}z$?

 

[Mark44]

I have $r=mz+b$ and $r(0)=0$, so $r=\frac{R}{h}z$?

If $r = mz + b$ then $r(0) \ne 0$.

 

[archaic]

If $r = mz + b$ then $r(0) \ne 0$.

The way I have drawn this is with $r(0)=0$, took two distinct points on the $z$ axis and told myself I am going to calculate the volume of that frustum.

 

[archaic]

 

[Mark44]

The way I have drawn this is with $r(0)=0$, took two distinct points on the $z$ axis and told myself I am going to calculate the volume of that frustum.

Do you want help or not? I am telling you that your line equation is wrong, and have told you why it is wrong.

On your 2nd drawing, label the points at the two ends of the slant line. Use them to find the equation of this line. Substitute in your integral, and integrate.

 

[archaic]

I have $\frac{R}{h}=\frac{r_2}{H}\Leftrightarrow H=\frac{hr_2}{R}$
$$f(H-h)=\frac{R}{h}(H-h)+b=r_1\Leftrightarrow \frac{R}{h}H-R-r_1=-b\Leftrightarrow r_2-R-r_1=-b=0\\
f(H)=\frac{R}{h}H+b=r_2\Leftrightarrow r_2-r_2=-b=0$$
The way you asked:
$$r-r_2=\frac{R}{h}(z-H)\Leftrightarrow r=\frac{R}{h}z-r_2+r_2=\frac{R}{h}z\\
r-r_1=\frac{R}{h}(z-(H-h))\Leftrightarrow r=\frac{R}{h}z-r_2+R+r_1=\frac{R}{h}z$$

 

[Mark44]

On second thought, your line equation is OK. I think your error comes from when you integrated.

I need to take off now, but will be back in a few hours.

 

[archaic]

On second thought, your line equation is OK. I think your error comes from when you integrated.

I need to take off now, but will be back in a few hours.

Thank you for your time, though!

 

[Mark44]

It might help to draw your picture with the vertical axis going up through the low point on the slant line. This will change your line equation.
Also, use r and R for the two radii of the frustum, but use y and x for the coordinate axes. In particular, don't use r for the small radius and also for the distance from the horizontal axis up to the slant line.

If you do the above, your integral will be $\int_{x = 0}^h \pi y^2 dx$, where y is the y-value on the slant line.

 

[archaic]

I think your error comes from when you integrated.

Indeed! What I have done was $\int_a^bx^2dx=\frac{(b-a)^3}{3}$, but it should have been $\frac{b^3-a^3}{3}$
Thank you for your help!

 

[Mark44]

Indeed! What I have done was $\int_a^bx^2dx=\frac{(b-a)^3}{3}$, but it should have been $\frac{b^3-a^3}{3}$
Thank you for your help!

You're welcome!