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volume of revolution question!

image-736431c0-1393-48e3-9087-02ee74cdba14-289273712-compressed.jpg.jpeg
Reply 1
Question 8, part ii)
Original post by Moltenmo
Question 8, part ii)


Remember that 1 litre = 1000 cm cubed

So, you want to find the volume inside the pot that has height of 45. Since the base starts at y=0.1102=10y=0.1 \cdot 10 ^2=10 then the upper limit will be at y=10+45=0.01x2y'=10+45=0.01 x^2 so x=522x=5 \sqrt{22}.

Once you find the volume, determine the amount of litres it converts to
Reply 3
Original post by RDKGames
Remember that 1 litre = 1000 cm cubed

So, you want to find the volume inside the pot that has height of 45. Since the base starts at y=0.1102=10y=0.1 \cdot 10 ^2=10 then the upper limit will be at y=10+45=0.01x2y'=10+45=0.01 x^2 so x=522x=5 \sqrt{22}.

Once you find the volume, determine the amount of litres it converts to

In doing this, I get:
image-0b5a8d72-034f-4160-bd12-9862f2bf2cd8419959082-compressed.jpg.jpeg
Which is wrong haha
(edited 6 years ago)
Original post by Moltenmo
In doing this, I get:

Which is wrong haha


I just noticed that you rotated it around the x-axis, not the y-axis.

For the y-axis, you want π1055x2.dy\pi \displaystyle \int_{10}^{55} x^2 .dy
Reply 5
Original post by RDKGames
So then it would be 44 litres of compost to get the the 45cm depth. If that's not the answer, then I must've missed something that I can't spot, someone else will be able to assist then.


I got it now but thank you anyways :biggrin:

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Reply 6
Original post by RDKGames
I just noticed that you rotated it around the x-axis, not the y-axis.

For the y-axis, you want π1055x2.dy\pi \displaystyle \int_{10}^{55} x^2 .dy


Yup, you're right.

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(edited 6 years ago)
Original post by Moltenmo
It should be 10x in place of x^2 but otherwise correct.

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Should be 10y10y :smile:
Reply 8
Original post by RDKGames
Should be 10y10y :smile:


Just noticed my idiocy right as you said that :smile:

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Reply 9
Being odd, we can also see the volume as:

(Volume)=1055π(10y)2dy \displaystyle \text{(Volume)} = \int^{55}_{10} \pi ( \sqrt{10y} )^{2} \, dy ?

(Also making sure the post is not overfilled).
(edited 6 years ago)
Reply 10
I suppose we can see it as that, but what good will it do?

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Reply 11
Since we are at it, 9iii) help please :biggrin:

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Original post by Moltenmo
Since we are at it, 9iii) help please :biggrin:

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Well... what's the volume of the vase? Then just times that by 232^3 before dividing it through by 1000
Reply 13
Original post by RDKGames
Well... what's the volume of the vase? Then just times that by 232^3 before dividing it through by 1000


Ahh, I see. Why are we multiplying it by 2^3 though?

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Original post by Moltenmo
Ahh, I see. Why are we multiplying it by 2^3 though?

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Because each unit is 2cm, so when you find the volume, you get units3\text{units}^3 but since unit=2cm\text{unit}=2 \text{cm} you just get 23cm32^3 \text{cm}^3

Might be easier to think of it as scaling the volume, and from GCSE you should know that volume scales by the cube of the scalar since a 3D object has 3 perpendicular axis along which the scalar takes effect.

It's like saying "OK, the unit cube has volume 1 unit cubed. If it's enlarged by 2 along every side, what's the new volume?" and so each side enlarges by two, hence the new vol is 222=1232 \cdot 2 \cdot 2=1 \cdot 2^3
(edited 6 years ago)
Reply 15
Original post by RDKGames
Because each unit is 2cm, so when you find the volume, you get units3\text{units}^3 but since unit=2cm\text{unit}=2 \text{cm} you just get 23cm32^3 \text{cm}^3

Might be easier to think of it as scaling the volume, and from GCSE you should know that volume scales by the cube of the scalar since a 3D object has 3 perpendicular axis along which the scalar takes effect.

It's like saying "OK, the unit cube has volume 1 unit cubed. If it's enlarged by 2 along every side, what's the new volume?" and so each side enlarges by two, hence the new vol is 222=1232 \cdot 2 \cdot 2=1 \cdot 2^3


Ahh, that makes sense. Thank you!

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