# Thread: Integration not necessary for volume question?

1. ## Integration not necessary for volume question?

Here is the question:

Determine the volume of a solid generated by rotating the triangle with vertices (0,0), (2,0), (1,1) about the line x = 2.

Now, I took the integral route, using the "washer" method (adding the "slices" from y = 0 to 1) and got an answer of 2*pi.

I also noticed that due to the simplicity of the triangle, we can interpret the distance from (0,0) to (1,1) and (1,1) to (2,0) to be sqrt(2) for each line segment. From that, the area of the triangle is [(sqrt(2))^2]/2 = 2/2 = 1

Then, looking at this 2-dimensional triangle "from above", we can multiply the area by 2*pi, resulting in a volume of 2*pi for the solid. A more or less obvious observation, I know, but I thought it might be nice to share this.

Also, if anyone can confirm that the volume is indeed 2*pi, I'd be very grateful!

2. Originally Posted by Tulki
Here is the question:

Determine the volume of a solid generated by rotating the triangle with vertices (0,0), (2,0), (1,1) about the line x = 2.

Now, I took the integral route, using the "washer" method (adding the "slices" from y = 0 to 1) and got an answer of 2*pi.

I also noticed that due to the simplicity of the triangle, we can interpret the distance from (0,0) to (1,1) and (1,1) to (2,0) to be sqrt(2) for each line segment. From that, the area of the triangle is [(sqrt(2))^2]/2 = 2/2 = 1

Then, looking at this 2-dimensional triangle "from above", we can multiply the area by 2*pi, resulting in a volume of 2*pi for the solid. A more or less obvious observation, I know, but I thought it might be nice to share this.

Also, if anyone can confirm that the volume is indeed 2*pi, I'd be very grateful!
The solid in question is a frustrum of a cone where a smaller cone was drilled out from the top. (See attachment)

With you example you know about the frustrum:
$r_1 = 2,~ r_2 = 1,~ h= 1$

Then the volume of the frustrum is:

$V_{fc} = \tfrac13 \cdot \pi\cdot h(r_1^2+r_2^2+r_1 r_2)$

The demensions of the cone are:

$r = 1,~ h=1$

The volume of the cone is

$V_{cone} = \tfrac13 \cdot \pi \cdot r^2 \cdot h$

Plug in these values to determine the volume.

The final result is indeed $V = 2\pi$

Originally Posted by Tulki
...

Then, looking at this 2-dimensional triangle "from above", we can multiply the area by 2*pi, resulting in a volume of 2*pi for the solid. A more or less obvious observation, I know, but I thought it might be nice to share this. <<<<<<<< You have used here (incorrectly!) Pappus 2nd theorem (or as I'm used to call it Guldin's rule). Your result is correct (by accident and by sheer luck)

...
Have a look here: Solid of revolution - Wikipedia, the free encyclopedia

4. Luckily I did not actually use Guldin's theorem for the question, and instead used the integral "washer" method to obtain the volume.