# Volume from revolved area?

• Nov 11th 2011, 06:11 AM
Bwts
Volume from revolved area?
Hi,

Hope I can explain this well. I have a function that gives me the deflection of a circular/disk membrane as a function of the radius. From this I can get the cross sectional area/profile of the disks displacement in terms of the radius. I would now like to revolve this area to create a volume.

My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis whilst I want to revolve it around the y-axis.

My inital attempt was to invert the function tbut this has proven impossible :(

My questions are:

1) Would the volume be same reguardless of which axis I choose to revolve around?

2) Is there an better way of doing this?

Attachment 22711
• Nov 11th 2011, 06:28 AM
Prove It
Re: Volume from revolved area?
Quote:

Originally Posted by Bwts
Hi,

Hope I can explain this well. I have a function that gives me the deflection of a circular/disk membrane as a function of the radius. From this I can get the cross sectional area/profile of the disks displacement in terms of the radius. I would now like to revolve this area to create a volume.

My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis whilst I want to revolve it around the y-axis.

My inital attempt was to invert the function tbut this has proven impossible :(

My questions are:

1) Would the volume be same reguardless of which axis I choose to revolve around?

2) Is there an better way of doing this?

Attachment 22711

If you want to revolve the region around the y axis, first write the function as a function of y, i.e. x = f(y).

Then, imagine the area being approximated using horizontal rectangular strips.

Each rectangular strip has length $\displaystyle x$ and width $\displaystyle \Delta y$, (some small change in y).

When you rotate the strips, they create cylinders, each of radius $\displaystyle x$ and depth $\displaystyle \Delta y$.

So the volume of each strip is $\displaystyle \pi x^2 \Delta y$, and therefore the volume of the region is approximated by

\displaystyle \begin{align*} V &\approx \sum{\pi x^2 \Delta y} \\ V &\approx \sum{\pi\left[f(y)\right]^2\Delta y} \end{align*}

As you increase the number of subdivisions, i.e. make $\displaystyle n \to \infty$ and $\displaystyle \Delta y \to 0$, this sum converges to an integral, and the approximation becomes exact.

So $\displaystyle V = \int_a^b{\pi \left[f(y)\right]^2\,dy}$, where $\displaystyle a, b$ are your y endpoints.
• Nov 11th 2011, 06:39 AM
Bwts
Re: Volume from revolved area?
Hi,

Thanks for your speedy response. My problem is getting x=f(y) see attched image.

evrything is constant except x. I keep getting a nasty looking polynomial in x which I cant reduce further

Attachment 22712
• Nov 11th 2011, 07:31 AM
tom@ballooncalculus
Re: Volume from revolved area?
Quote:

Originally Posted by Bwts
1) Would the volume be same reguardless of which axis I choose to revolve around?

No. Try to imagine the two.

Quote:

Originally Posted by Bwts
2) Is there a better way of doing this?

Yes, integrate along the x axis and rotate the vertical strips around the y axis to make hollow cylinders.

The formula is then

$\displaystyle V = \int_a^b{2\pi x f(x) dx$

... as explained admirably at Solid of revolution - Wikipedia, the free encyclopedia. I thought that page had a typo, but I was misreading 'revolving an area between curve and x-axis' as 'revolving around the x axis'. Maybe you did, too...

Quote:

Originally Posted by Bwts
My problem is that when ever I look this up on the internet I find equations for the volume if I revolve the area aournd the x-axis

• Nov 11th 2011, 07:39 AM
Bwts
Re: Volume from revolved area?
Hi Tom,

Just so I'm clear. I should intergrate my function along x then multiply this intergrated function by the original function and intergrate the product between my x limits?
• Nov 11th 2011, 07:48 AM
tom@ballooncalculus
Re: Volume from revolved area?
Integrate with respect to x, yes, and between your x limits, yes... not sure I follow your multiplying proposal... what you should be planning, anyway, is to integrate the product which is the result of multiplying your function (the height of each hollow cyclinder) by 2 pi x (the circumference of the same cylinder).
• Nov 11th 2011, 08:19 AM
Bwts
Re: Volume from revolved area?
Ahh yes I get you now...thanks Tom