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Volume of rotation: x=(y-10)^2, x=1, rotated about y=9  TOPIC_SOLVED

(1) The region bounded by the given curves is rotated about y = 9. x=(y-10)^2 , x=1. Find the volume V of the resulting solid by any method. and (2) The region bounded by the given curves is rotated about the y-axis. y=-x^2+15x-54, y=0. Find the volume V of the resulting solid by any method. I got (...
by chinex9a
on Mon Sep 13, 2010 4:46 am
 
Forum: Calculus
Topic: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9
Replies: 5
Views: 3527

Re: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9  TOPIC_SOLVED

I integrated from 11 to 9 pi*((y-10)^2)^2 dy
I know its not correct but wot did u get and hw did u do it. I did not know wot to do with the x=1 line
by chinex9a
on Mon Sep 13, 2010 12:09 pm
 
Forum: Calculus
Topic: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9
Replies: 5
Views: 3527

Re: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9  TOPIC_SOLVED

I actually got that at some point but min was a negative anwser and why is it (y-9) and not (9-y).

Thank You for showing me how to do it. I usually get confused when you are meant to rotate functions about lines other than the y-axis and x-axis.
by chinex9a
on Mon Sep 13, 2010 3:45 pm
 
Forum: Calculus
Topic: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9
Replies: 5
Views: 3527

how to integrate x^2/(36-x^2)^(3/2): parts or substitution?  TOPIC_SOLVED

How would you evaluate this integral ?

x^2/(36-x^2)^(3/2)

Parts or Substitution ?(I tried using substitution) Please explain.
by chinex9a
on Fri Sep 24, 2010 9:21 pm
 
Forum: Calculus
Topic: how to integrate x^2/(36-x^2)^(3/2): parts or substitution?
Replies: 1
Views: 1310

Partial Fraction integration  TOPIC_SOLVED

Integrate sqrt(x)/(x^2+x) from 1/3 to 3

How do you go about the integration
by chinex9a
on Sat Oct 09, 2010 8:06 pm
 
Forum: Calculus
Topic: Partial Fraction integration
Replies: 2
Views: 1263

Volume of solid under the curve y= 7/(x^2 +5x +6)  TOPIC_SOLVED

I am studying for an exam and I don't you know how to go about this? Volume of solid under the curve y= 7/(x^2 +5x +6) from x = 0 to x = 1 if rotated about the y-axis. I made x the subject of the formula getting (sqrt(y+28)-5 sqrt(y))/(2 sqrt(y)) and then I found what y is when x= 0 and x=1, which w...
by chinex9a
on Sat Oct 09, 2010 10:50 pm
 
Forum: Calculus
Topic: Volume of solid under the curve y= 7/(x^2 +5x +6)
Replies: 8
Views: 2589

Re: Volume of solid under the curve y= 7/(x^2 +5x +6)  TOPIC_SOLVED

because we are rotating about the y axis :confused:
by chinex9a
on Sun Oct 10, 2010 3:23 am
 
Forum: Calculus
Topic: Volume of solid under the curve y= 7/(x^2 +5x +6)
Replies: 8
Views: 2589

Re: Volume of solid under the curve y= 7/(x^2 +5x +6)  TOPIC_SOLVED

Yes that is what I did right. pi*(f(y))^2 dy
by chinex9a
on Sun Oct 10, 2010 4:41 am
 
Forum: Calculus
Topic: Volume of solid under the curve y= 7/(x^2 +5x +6)
Replies: 8
Views: 2589

Re: Volume of solid under the curve y= 7/(x^2 +5x +6)  TOPIC_SOLVED

and so are b and a 7/6 and 7/12 ?
by chinex9a
on Sun Oct 10, 2010 5:13 am
 
Forum: Calculus
Topic: Volume of solid under the curve y= 7/(x^2 +5x +6)
Replies: 8
Views: 2589

Re: Volume of solid under the curve y= 7/(x^2 +5x +6)  TOPIC_SOLVED

Am I correct???????????? :?:
by chinex9a
on Mon Oct 11, 2010 3:54 pm
 
Forum: Calculus
Topic: Volume of solid under the curve y= 7/(x^2 +5x +6)
Replies: 8
Views: 2589

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