Calculating Volume with Volumes of Revolution: Triangle Rotated about x = -2

[Panphobia]

Homework Statement

Calculate the volume obtained by rotating the triangle bounded by y = 0, y = x, and y = 2 - x, about the line x = -2. You may use either horizontal or vertical rectangles.

The Attempt at a Solution

So since this is a triangle, I tried to split up the volume down to two integrals, one from 0 - 1, and the other from 1-2. I used vertical rectangles by the way. So I set up my first integral as

2 \cdot \pi \int_{0}^{1} (x+2)(x) \,dx
since the shell radius would be a distance x + another 2, and then the shell height would be the x.
My second integral is

2 \cdot \pi \int_{1}^{2} (x+2)(2-x) \,dx

then I added these together, but I did not get the correct answer, could anyone show me what the error in my steps are?

 

[BvU]

I am issing 2). relevant equations. Where did you get the idea to set up the integrals as you did ?
Did you make a drawing to see if you want to integrate over dx or over dy ?

[edit]Ok, you are using vertical shells. What is their height ? x seems strange: in my drawing height decreases with x.

[edit][edit]vertical shells require bounds in x, not in y.

 

[Panphobia]

y=x is the first height,no? from 0-1 if not, then what is it

 

[BvU]

Ah, I see. You want to integrate over x (i.e. horizontal rings), the bounds are x=0 at the lower en x=x at the upper end , so the height is x for the first integral. And also for the second integral.

Or do you want to use vertical rings and integrate over x ? x does not exceed 1, right ?

Can you show your drawing ?

 

[Panphobia]

I vertical cylinders, I don't think you can have vertical rings. You can have horizontal rings. Here is the graph

 

[LCKurtz]

Homework Statement

Calculate the volume obtained by rotating the triangle bounded by y = 0, y = x, and y = 2 - x, about the line x = -2. You may use either horizontal or vertical rectangles.

The Attempt at a Solution

So since this is a triangle, I tried to split up the volume down to two integrals, one from 0 - 1, and the other from 1-2. I used vertical rectangles by the way. So I set up my first integral as

2 \cdot \pi \int_{0}^{1} (x+2)(x) \,dx
since the shell radius would be a distance x + another 2, and then the shell height would be the x.
My second integral is

2 \cdot \pi \int_{1}^{2} (x+2)(2-x) \,dx

then I added these together, but I did not get the correct answer, could anyone show me what the error in my steps are?

Your integrals are set up correctly. You must have some mistake in your evaluating them.

 

[Panphobia]

Yea, I think I did miss a factor of (1/3) somewhere when I solved my integral. Thanks for the help!

 

[BvU]

Oh boy, my mistake: I revolved around x= -2 and had line x=0 as boundary instead of y=0. Scuse me! Eating humble pie. At least it explains why communicating was difficult.

And yes, integrals are set up correctly -- from the start .