# for those who want P2 practice

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#1
i've just done this beast of a P2 volumes of revolution question.

try it, it's good practice.

The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis

a)12pi b)206pi/15
0
15 years ago
#2
(Original post by hihihihi)
i've just done this beast of a P2 volumes of revolution question.

try it, it's good practice.

The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis

a)12pi b)206pi/15
a. volume= 20pi-pi*int(x-1)[5 and 1]
=20pi-pi(x^2/2-x)
=20pi-pi(12.5-5-0.5+1)
=20pi-8pi
=12pi
0
15 years ago
#3
0 and 1 i think
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15 years ago
#4
(Original post by lgs98jonee)
wot r the limits when u integrate around the xaxis??
well you know y = 2...

cmon dude, dont slip now!
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15 years ago
#5
oh yeh lol i fell in the same trap i didnt see y=2! deary me
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#6
(Original post by lgs98jonee)
a. area= 6-pi*int(x-1)[3 and 1]
=6-pi(x^2/2-x)
=6-pi(4.5-0.5-3+1)
=6-2pi
area?? it's volumes!!
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15 years ago
#7
yeh i just noticed the y=2! oops
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15 years ago
#8
(Original post by hihihihi)
you also need to find another limit..clue: when y=2, x=...
oh thnx for that clue.. .give us a chance to do the question....seems like a really patronisng tone..ok then children wot if y=1+x and x=1, y=...
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15 years ago
#9
r u sure that it is not 12.5pi?
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15 years ago
#10
I got for a) (25pi-5pi)-(1pi-1pi) = 20pi and for b) 404/15pi...
When y = 2... 4 = x-1 so x=5... You seem to have integrated the upper value as being 4... Why?
For a) y was squared already so I just integrated x^2 - x between 5 and 1...

Have I done something horrible wrong?
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15 years ago
#11
why do u do 20pi-what u get? where are u minusing it from?
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15 years ago
#12
I am such a retard... I forgot to find the intersection points... I will fail this bloody exam!
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15 years ago
#13
(Original post by hihihihi)
i've just done this beast of a P2 volumes of revolution question.

try it, it's good practice.

The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis

a)12pi b)206pi/15

b. volume=piINT(y^2+1)^2 dx
=pi INT(y^4+2y^2+1)
=pi(y^5/5+2y^3/3+x) [between 2 and 0]
=pi(32/5+16/3+2)
=96/15+80/15+30/15
=206pi/15
0
15 years ago
#14
(Original post by Sapphira)
why do u do 20pi-what u get? where are u minusing it from?
well i did the big cylinder (pi *2^2*5)-the area under the curve between x=0 and x=5
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#15
(Original post by lgs98jonee)
a. volume= 20pi-pi*int(x-1)[5 and 0]
=20pi-pi(x^2/2-x)
=20pi-pi(12.5-5)
=20pi-7.5pi
=12.5pi
you need to do between 5 and 1??
0
#16
(Original post by lgs98jonee)
b. volume=piINT(y^2+1)^2 dx
=pi INT(y^4+2y^2+1)
=pi(y^5/5+2y^3/3+x) [between 2 and 0]
=pi(32/5+16/3+2)
=96/15+80/15+30/15
=206pi/15
high five!!
0
15 years ago
#17
(Original post by hihihihi)
high five!!
they werent easy
0
15 years ago
#18
The diagram shows the region R bounded by the curve y^2=x-1, the x-axis and y-axis, and the line y=2

Find the volume when R is rotated 360 degrees around:
a) the x axis
b) y axis

a)

2^2 = x-1
x = 5

Vol. = (5/0){ pi(y^2) dx
= (5/0){ pi x - pi) dx
= [pi.x^2/2 - pi.x] (5/0)
= [pi.25/2 - 5pi] - 0
= 12.5pi

b)
y^2=x-1
x = (y^2 + 1)
x^2 = y^4 + 2y^2 + 1

Vol. = (2/0){ pi(x^2) dy
= (2/0){ (pi.y^4 + 2pi.y^2 + pi) dy
= [pi.y^5/5 + 2pi.y^3/3 + pi.y](2/0)
= [(32/5)pi + (16/3)pi + (2)pi]
= (206/15)pi

0
15 years ago
#19
(Original post by mik1a)

Vol. = (5/0){ pi(y^2) dx
= (5/0){ pi x - pi) dx
= [pi.x^2/2 - pi.x] (5/0)
= [pi.25/2 - 5pi] - 0
= 12.5pi
i got that but apparently it was 12pi :-(
0
15 years ago
#20
I don't get why between 5 and 1.
0
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