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

Limits, differentiation, related rates, integration, trig integrals, etc.
chinex9a
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Volume of rotation: x=(y-10)^2, x=1, rotated about y=9

Postby chinex9a » Mon Sep 13, 2010 4:46 am

(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 (2*pi)/5 for number 1 and (135*pi)/2 for second one, dunno if it is correct.

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

Postby Martingale » Mon Sep 13, 2010 11:53 am

chinex9a wrote:(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 (2*pi)/5 for number 1 and (135*pi)/2 for second one, dunno if it is correct.


Just looking at the first one, that is not the answer I got. Can you show us what you did?

chinex9a
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Re: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9

Postby chinex9a » Mon Sep 13, 2010 12:09 pm

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

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

Postby Martingale » Mon Sep 13, 2010 12:58 pm

chinex9a wrote: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





or


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

Postby chinex9a » Mon Sep 13, 2010 3:45 pm

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.

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

Postby Martingale » Mon Sep 13, 2010 4:26 pm

chinex9a wrote: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.


on the bounded region y is larger than 9. You want a positive volume so we use y-9


the usual formula is




for , is the height of the line you are rotating about


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