Core Maths 3: Integration to find volume of revolution ii
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Core Maths 3: Integration to find volume of revolution iiHi 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?? -
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Re: Core Maths 3: Integration to find volume of revolution iiRequired volume(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??
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Re: Core Maths 3: Integration to find volume of revolution iiUse the reverse bracket rule when integrating and you should end up with 18 -0 when you sub in the 8 and 2 hence 18pi.(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??
I just tried it out on a piece of paper and got the right answer.Last edited by TLK; 05-04-2012 at 20:01. -
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Re: Core Maths 3: Integration to find volume of revolution ii(Original post by TenOfThem)
32pi is wrong for your cylinder ... I think you have used 8 instead of 6
I did use 8, I didn't use another limit as I assumed it was 0 -
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Re: Core Maths 3: Integration to find volume of revolution iiFirstly ... just looking at the diagram you should have realised that the other limit was 2(Original post by jackie11)
I did use 8, I didn't use another limit as I assumed it was 0
Secondly ... even when a limit is 0 you cannot just ignore it as the value may not always be 0 -
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Re: Core Maths 3: Integration to find volume of revolution iiIntegration here takes the same time as using the area of cylinder takes, so i prefer to use integration rather than(Original post by TenOfThem)
I find it odd that both the OP and yourself would use integration to find the volume of a cylinder
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Re: Core Maths 3: Integration to find volume of revolution iireally?(Original post by raheem94)
Integration here takes the same time as using the area of cylinder takes [/latex]
pi x 4 x 6 does not really take any time -
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Re: Core Maths 3: Integration to find volume of revolution iiStill the probability of error in such a question will be very less. This is basic integration.(Original post by TenOfThem)
Much more scope for error though ... not saying that you are wrong ... just that I find it odd -
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Re: Core Maths 3: Integration to find volume of revolution iiGiven the errors made by the OP I beg to differ(Original post by raheem94)
Still the probability of error in such a question will be very less. This is basic integration. -
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Re: Core Maths 3: Integration to find volume of revolution iiActually 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.(Original post by TenOfThem)
Given the errors made by the OP I beg to differ
![\displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi](http://www.thestudentroom.co.uk/latexrender/pictures/98/98ce8565a0b52b5958ef84b427d6ad7d.png "\displaystyle \pi \int^8_2 4dx = \pi \left[4x\right]^8_2=\pi(32-8)=24\pi")
