## 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

(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.

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

(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

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

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
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### Re: Volume of rotation: x=(y-10)^2, x=1, rotated about y=9

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
$\int\limits_{9}^{11}2\pi(y-9)(1-(y-10)^2)dy=\frac{8\pi}{3}$

or

$\int\limits_{0}^{1}\pi\left((\sqrt{x}+10)-9\right)^2dx-\int\limits_{0}^{1}\pi\left((-\sqrt{x}+10)-9\right)^2dx=\frac{8\pi}{3}$

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

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.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

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

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

$\int\limits_{a}^{b}2\pi(z-r)f(z)dz$

for $r\leq a$, $r$ is the height of the line you are rotating about