Another question!! Vol of Revolution Watch

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appleshampoo
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#1
Report Thread starter 13 years ago
#1
ok im stuck again!!

Curve given by eq's x=2t, y=t^2
(find eq of normal at P where t=3, found this is 3y+x=33, or y=(-1/3)x+11)

Then it says R is bound by the curve btwn the origin and P and the line OB and BP (B being where the normal meets the y-axis - i think this is 11)

R is rotated through 2pie about y axis - find the vol (answer is 186pie cubic units)

so i tried but got wrong answer:
(let I = integral)
[pie I (33-3y)^2 dy] - [pie I 4y] (with y limits of 11 and 0) ???????

Thanks!!!
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mc_watson87
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#2
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Hmmm I got 4044 pi =\ lol

*starts again*

Edit: You and I went wrong cus we used the equation of the normal in the integration rather than the equation of the curve.
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Chiral
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#3
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Oh I love V of R

OK, you've got the curve in parametric form. And we are turning it about the Y axis.

I'll have to use 'I' for an integral sign

So we need pi^2 I(x^2)dy

first we need the cartesian equation of the curve.

x=2t, y=t^2
t = root y
x = 2 root y

then we integrate

pi I x^2 dy
= pi I (2 root y)^2 dy
= pi I 4y dy
= pi [2y^2]

now you just need to substitue in your _Y_ limits. Which you seem to say is 11

that would make it 242 pi...

I'm fairly certain my integration is right, so the limits might be wrong, I didn't bother doing the differentiation. Can you scan and post the the original question?
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mc_watson87
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Use limits of 0 to 9 above and then add 1/3 r^2 h and that gets u the 186

Cus its actually bounded to that amount of the x limit and then u add the volume of a cone for the bit above which is (1/3 * 6^2 * 2)Pi
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dvs
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#5
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Eliminate t:
x² = 4y

Volume of the curve betewen 0 and y-coord of P (y=9) = 2 pi ∫ x² dy
= pi ∫ 4y dy
= 2pi [y²]{0, 9}
= 162pi

Volume of the cone on top of the curve = (1/3) pi r² h
= (1/3) pi (x-coord of P)² (B - y-coord of P)
= (1/3) pi * 6² * (11-9)
= 24pi

Total volume = 24pi + 162pi = 186pi
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Chiral
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dah, I get it now. ASCII text is rubbish for maths.
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appleshampoo
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#7
Thank you so much...but.....

how do u know y-coord of P is 9?

why is it the vol of a cone on top of the curve? why dont you work out volume of the normal rotated minus the volume of the curve rotated to get the volume of R bounded in between???

So confused!!
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dvs
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Find the coords of P by plugging t=3 into x and y.

[See attachment.]
The blue region is the volume of the curve rotated about the y-axis between the points y=0 and y=9.

The red region is a triangle, and when you rotate it about the y-axis it becomes a cone. The radius of the cone is the base of the triange and its altidude is the same as the triangle's.
Attached files
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appleshampoo
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#9
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oh i see! i was just looking at the graph in the wrong way1 i dint think about splitting R into two sections!

Thanks a lot!!!
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Chiral
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#10
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#10
dvs...NOW I understand what the damn question was about!
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