# Thread: Volumes of solid of revolution, rotated about the y axis

1. ## Volumes of solid of revolution, rotated about the y axis

I am not sure how to tackle this one.

find the volume of y=x+1: x=0 and x=2, rotated about the y axis.

2. ## Re: Volumes of solid of revolution, rotated about the y axis

Originally Posted by johnsy123
I am not sure how to tackle this one.

find the volume of y=x+1: x=0 and x=2, rotated about the y axis.
Possibly that is not the exact formulation of the problem, the volume is not well defined.

3. ## Re: Volumes of solid of revolution, rotated about the y axis

The previous question to this i had to work out the volume rotated about the x axis, this is where i got the points x=0 and x=2 from, but now they want me to find the volume of the solid about the yaxis. Should i sub the two x numbers into the function, to obtain y values? than work from there.

4. ## Re: Volumes of solid of revolution, rotated about the y axis

I think that's a good solution, you have also to express now $y$ in function of $x$, so: $x=y-1$ and then use the formula (in general):
$V=\pi \int_{a}^{b} [f(y)]^2dy$

5. ## Re: Volumes of solid of revolution, rotated about the y axis

Originally Posted by johnsy123
I am not sure how to tackle this one.

find the volume of y=x+1: x=0 and x=2, rotated about the y axis.
as stated by Fernando, the three given lines do not enclose a region in the plane.

however, if you're speaking of the region in quadrant I bounded by the given lines as shown in the graph ...

$V = \pi \int_0^2 (x+1)^2 \, dx$

a. using the method of cylindrical shells w/r to x

$V = 2\pi \int_0^2 x(x+1) \, dx$

b. using the methods of disks and washers w/r to y (requires two integrals)

$V = \pi \int_0^1 2^2 \, dy + \pi \int_1^3 2^2 - (y-1)^2 \, dy$

c. finally, and probably the easiest, using geometry (cylinder volume - cone volume)

$V = \pi \cdot 2^2 \cdot 3 - \frac{\pi}{3} \cdot 2^2 \cdot 2$

6. ## Re: Volumes of solid of revolution, rotated about the y axis

Why do you have to multiply by 2pi and not pi in previous cases?

7. ## Re: Volumes of solid of revolution, rotated about the y axis

Originally Posted by johnsy123
Why do you have to multiply by 2pi and not pi in previous cases?
Skeeter said he was using cylindrical shells. The surface area of a cylinder of radius r and height h is $2\pi rh$, the circumference of the cylinder times the height of the cylinder.