# for those who want P2 practice

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

try it, it's good practice.

answers:

a)12pi b)206pi/15

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**answers:

a)12pi b)206pi/15

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#2

(Original post by

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

try it, it's good practice.

answers:

a)12pi b)206pi/15

**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**answers:

a)12pi b)206pi/15

=20pi-pi(x^2/2-x)

=20pi-pi(12.5-5-0.5+1)

=20pi-8pi

=12pi

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#4

(Original post by

wot r the limits when u integrate around the xaxis??

**lgs98jonee**)wot r the limits when u integrate around the xaxis??

cmon dude, dont slip now!

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(Original post by

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

**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

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#8

(Original post by

you also need to find another limit..clue: when y=2, x=...

**hihihihi**)you also need to find another limit..clue: when y=2, x=...

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#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?

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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#12

I am such a retard... I forgot to find the intersection points... I will fail this bloody exam!

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#13

**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**

answers:

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

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#14

(Original post by

why do u do 20pi-what u get? where are u minusing it from?

**Sapphira**)why do u do 20pi-what u get? where are u minusing it from?

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(Original post by

a. volume= 20pi-pi*int(x-1)[

=20pi-pi(x^2/2-x)

=20pi-pi(12.5-5)

=20pi-7.5pi

=12.5pi

**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

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(Original post by

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

**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

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#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

Nice question, had to think hard about the (b) part!

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#19

(Original post by

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

**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

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