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Integration and Homebrewing.

Ok so I've been thinking of practical ways I can use integration at home and have decided to check the volume of one of my homebrewing buckets is correct.

I know how to find the volume using integration, but how do I find the volume of this specific bucket without a function for it?

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Original post by Banny Dyrne
Ok so I've been thinking of practical ways I can use integration at home and have decided to check the volume of one of my homebrewing buckets is correct.

I know how to find the volume using integration, but how do I find the volume of this specific bucket without a function for it?


You'll need to create a function.

What's the exact shape? And any pertinent dimensions?
Reply 2
Its just an average bucket shape.

Top diameter is 35.2cm, bottom is 30.6cm and its 43cm tall.

I can't remember much of GCSE maths (too long ago and didn't listen) but there is probably a way of finding out the volume just using these three measurements.

But since I cant do that I'm graphing this and integrating!
Original post by Banny Dyrne
Its just an average bucket shape.

Top diameter is 35.2cm, bottom is 30.6cm and its 43cm tall.

I can't remember much of GCSE maths (too long ago and didn't listen) but there is probably a way of finding out the volume just using these three measurements.

But since I cant do that I'm graphing this and integrating!


Well if we make the vertical axis of the bucket, the x-axis.

Then we have x runs from 0 to 43.

Our graph will be the radius - a straight line from 30.6/2 to 35.2/2

You just need to create a formula for it. Post if you need help on that.

And then do your integration.
Reply 4
I would model it as a volume of revolution, probably with a linear function if the bucket looks a lot like it has straight sides (which I think buckets usually do).
Reply 5
Original post by ghostwalker
Post if you need help on that.


Yes I do need help with that! Got almost no idea how to create a formula.


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Reply 6
Original post by Smaug123
I would model it as a volume of revolution, probably with a linear function if the bucket looks a lot like it has straight sides (which I think buckets usually do).


Yeah it does have straight sides. I am trying to model it as a volume of revolution but I don't know how to create a formula for the bucket. Everything else I can do.


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Reply 7
Estimate it. If your bucket is like mine then it's straight sides and ever so slightly wider at the top than the bottom, so find the gradient of the sides with a few measurements and simple geometry.
A carboy might be a bit more difficult, as it's curved and you can't put a tape measure in it.
Reply 8
But I only know how to find the gradient of sometime using coordinates. And I don't have coordinates. If its simple geometry I need I won't know what to do because I paid zero attention at school so I only know what I've taught myself, which is difficult geometry.


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Reply 9
So the difference in diameter increases from the bottom to the top at 4.6cm per 43cm. Don't know if I'm on the right track but surely that's useful


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Reply 10
I've ended up with y=1.069764x


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Reply 11
Original post by Banny Dyrne
I've ended up with y=1.069764x


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Don't forget a constant, so the radius when x = 0
Reply 12
Original post by Banny Dyrne
I've ended up with y=1.069764x

I'll put the origin at the centre of the base of the bucket, and look at it side-on. Then the edge starts at (0, 15.3) since the radius at the base is 15.3; the edge ends at (43,17.6) since it's 43cm tall and has radius 17.6 at the top.
Hence the line representing the edge is the line (17.6-15.3)/(43-0) = (y-15.3)/(x-0), or y = 15.3+0.0535x.
This needs to be rotated around the y-axis. Alternatively, you could invert it (represent x in terms of y) and then you could rotate around the x-axis.
Reply 13
Original post by Smaug123
I'll put the origin at the centre of the base of the bucket, and look at it side-on. Then the edge starts at (0, 15.3) since the radius at the base is 15.3; the edge ends at (43,17.6) since it's 43cm tall and has radius 17.6 at the top.
Hence the line representing the edge is the line (17.6-15.3)/(43-0) = (y-15.3)/(x-0), or y = 15.3+0.0535x.
This needs to be rotated around the y-axis. Alternatively, you could invert it (represent x in terms of y) and then you could rotate around the x-axis.


Yes! I just worked out x is equal to 0.0534883721 so I'm right up to that point. Thanks everyone for the help! Finally something useful I can personally do with integration!


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Original post by Banny Dyrne

Finally something useful I can personally do with integration!


Not only can you work out the volume of the bucket, but as a further challenge, you can create a scale to go on the side with depths for various volumes.
Reply 15
Done everything and it worked out at 36.614 litres which looks bang on. Cheers for the help everyone!


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I think this is how maths should be taught in school

Using it to find out everyday things
Reply 17
Original post by upthegunners
I think this is how maths should be taught in school
Using it to find out everyday things

Certainly I think that's how it should be introduced - then you can spot patterns and so on that you can use to make it more rigorous.
Original post by Smaug123
Certainly I think that's how it should be introduced - then you can spot patterns and so on that you can use to make it more rigorous.


There are people in my school who do Maths Alevel who wouldn't even know what the purpose of integral calculus is, they just know that you 'add one to the power and divide by the new power' and that is it.

Nothing bugs me more than people just applying methods without knowing what they are actually doing. I think before the Alevel exams are made 'tougher' a new approach to teaching would need to be considered within mathematics

Just my opinion :smile:
Reply 19
Original post by upthegunners
I think this is how maths should be taught in school

Using it to find out everyday things


They do in OCR MEI.

Just saying.

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