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Core Maths 3: Integration to find volume of revolution ii

Hi there

Please could you help me with this one:



I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

I don't really know how to work out the shaded area, any ideas??
Reply 1
32pi is wrong for your cylinder ... I think you have used 8 instead of 6
Reply 2
The curve is wrong too

Not sure how you have got the 2pi
Reply 3
When you integrated you say you used 8

What was your other limit
Reply 4
Original post by jackie11
Hi there

Please could you help me with this one:

I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

I don't really know how to work out the shaded area, any ideas??


Required volume =π(2822dx28(4x)2dx) \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)
Reply 5
Original post by jackie11
Hi there

Please could you help me with this one:

I have tried integrating both the line and the curve for the value of x = 8, and I am getting 2 pi for the area under the curve and 32 pi for the area under the line y = 2, the answer I have in my book for the whole thing is 18 pi.

I don't really know how to work out the shaded area, any ideas??


Use the reverse bracket rule when integrating and you should end up with 18 -0 when you sub in the 8 and 2 hence 18pi.

I just tried it out on a piece of paper and got the right answer.
(edited 11 years ago)
Reply 6
Original post by raheem94
Required volume =π(2822dx28(4x)2dx) \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)


I find it odd that both the OP and yourself would use integration to find the volume of a cylinder
Reply 7
Original post by TenOfThem
32pi is wrong for your cylinder ... I think you have used 8 instead of 6



Original post by TenOfThem
The curve is wrong too

Not sure how you have got the 2pi



Original post by TenOfThem
When you integrated you say you used 8

What was your other limit


I did use 8, I didn't use another limit as I assumed it was 0
Reply 8
Original post by raheem94
Required volume =π(2822dx28(4x)2dx) \displaystyle = \pi\left(\int^8_2 2^2dx - \int^8_2 \left(\frac4{x}\right)^2dx \right)


aww thank you, this is great!!!
Reply 9
Original post by jackie11
I did use 8, I didn't use another limit as I assumed it was 0


Firstly ... just looking at the diagram you should have realised that the other limit was 2

Secondly ... even when a limit is 0 you cannot just ignore it as the value may not always be 0
Reply 10
Original post by TenOfThem
I find it odd that both the OP and yourself would use integration to find the volume of a cylinder


Integration here takes the same time as using the area of cylinder takes, so i prefer to use integration rather than πr2h \displaystyle \pi r^2h
Original post by raheem94
Integration here takes the same time as using the area of cylinder takes


really?

pi x 4 x 6 does not really take any time
Reply 12
Original post by TenOfThem
really?

pi x 4 x 6 does not really take any time


π284dx=π[4x]28=π(328)=24π \displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi

The above will also not take much time, may be just 5 seconds more.
Original post by raheem94
π284dx=π[4x]28=π(328)=24π \displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi

The above will also not take much time, may be just 5 seconds more.


Much more scope for error though ... not saying that you are wrong ... just that I find it odd
Reply 14
Original post by TenOfThem
Much more scope for error though ... not saying that you are wrong ... just that I find it odd


Still the probability of error in such a question will be very less. This is basic integration.
Original post by raheem94
Still the probability of error in such a question will be very less. This is basic integration.


Given the errors made by the OP I beg to differ
Reply 16
Original post by TenOfThem
Given the errors made by the OP I beg to differ


Actually i was indicating my case not the OP's. I have seen several simple integration mistakes made by the OP so its best for the OP to avoid integration as much as possible.

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