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Integration (1 Viewer)

lyounamu

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FDownes said:
D'oh! For some reason I was substituting y = 0 instead of x = 0. Thanks. :)

EDIT: I think you may have made a mistake. Did you square x before integrating it?
No. I will fix it.

Can you wait for a bit? I have got a business to attend in few minutes. I will make sure I come back.
 
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FDownes

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Okay, thanks very much. :)

EDIT: BTW, the answer is 2pi/5, if you were wondering.
 

conics2008

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Doctor Jolly said:
All your working out scares me .. I've just started on Integration :cold:
hey his working out is step by step. I would say I skip alot of working out.

for 3units integration i dont need more than 7 lines ..

for 4units 10 lines is more then enough.

try to do them as fast as possible durning exam condition.
 

Doctor Jolly

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conics2008 said:
hey his working out is step by step. I would say I skip alot of working out.

for 3units integration i dont need more than 7 lines ..

for 4units 10 lines is more then enough.

try to do them as fast as possible durning exam condition.
I'm not ready to take such a quantum leap yet. I better just stick to the very very long ways of integration until I master it, before I confuse myself and end up cursing Newton. :)

EDIT: I like the integration sign .. it looks like a noodle hanging off the equation :p
 
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conics2008

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dolly dw integration aint that hard if you play by its rules.
 

FDownes

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Has anyone got an answer to the question? I'm still stuck... :(
 

conics2008

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hey im assuming your talking about the one up there.

solve the first bits

y= 2+2rootx

x=0 >> y =2
y=4 >> x=1

now find area by chaning the subject to x because its with the y axis

therefore y=2+2rootx written with sub x is x= 1/4 (y-2)^2

now u said rotated hence find x^2

x^2 = 1/16 ( y-2)^2 dy

pi/16 S between 4 and 2 (y-2)^4

integrate

you shouuld get pi/80 { (y-2 )^5 } between 4 and 2

sub in limits you get pi/80 * 32 ==== 2pi/5
 

FDownes

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Whoops, made a mistake when I was looking at the answers, so I thought the solution was 2pi/3. Geez, I can't even read the answers correctly. :rolleyes:
 

FDownes

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One more time;

The region bounded by the curve y = 2√(x) - 1, the lines x = 3, x = 4 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

Here's my working...

= y2 = (2x1/2 - 1)2
= y2 = 4x - 4x1/2 + 1
= V = pi 34∫ (4x - 4x1/2 + 1)
= V = pi[2x2 - (8x3/2)/3 + x]34
= V = [(2(4)2 - (8(4)3/2)/3 + (4)) - (2(3)2 - (8(3)3/2)/3 + (3))]
 
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tommykins

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FDownes said:
One more time;

The region bounded by the curve y = 2√(x) - 1, the lines x = 3, x = 4 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

Here's my working...

= y2 =
i'm assuming it's y = 2sqrt(x) - 1 sqrtx only.

y² = (2sqrtx -1)² = 4x - 4sqrtx + 1

V = pi int 4x - 4sqrtx + 1 from x = 4->3
= pi [2x² - [8x^(3/2)]/3 + x ] 4->3
~ 7.523 pi
 

Aerath

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I got (8r3 - 19/3)pi (which equals about 7.5pi).

You started off well with
y2 = 4x-4x0.5 + 1

Sub it into integration:
V = pi* J43 (4x-4x0.5 + 1)dx
= [2x2 - 8/3*x2/3 +x]34

Sub in the numbers, and you should get the answer.
 

FDownes

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The answer on the worksheet is pi(4√(3) - 19/3) units3. I didn't get this answer when subbing in the numbers...
 

Aerath

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FDownes said:
The answer on the worksheet is pi(4√(3) - 19/3) units3. I didn't get this answer when subbing in the numbers...
Neither did I. =\
 

tommykins

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FDownes said:
The answer on the worksheet is pi(4√(3) - 19/3) units3. I didn't get this answer when subbing in the numbers...
The answer should be pi(8√(3) - 19/3).
 

FDownes

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Hmmm... My working turns out quite close to the answer, maybe the solutions are wrong...

EDIT:

tommykins said:
The answer should be pi(8√(3) - 19/3).
Yeah, that's what I got.
 

Aerath

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Yeah, I'm pretty sure the answer's wrong.
 

conics2008

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y=2rootx-1

rotated at x gives you a disc with radius y. Hence volume of dics with width dx

piy^2 dx

y^2= 4x -4rootx +1

pi S between 4 and 3 4x -4rootx +1

= pi { 2x^2-8/3x^3/2 +x } 4 and 3
= pi/3 [ 24root3 - 19 ] cm^3
 

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