1. ## Volume (disk method)

find the volume of the solid genereated b revolving the triangular region bounded by the lines:

y = 2x , y = 0 , and x = 1 about

a. the line x = 1
b. the line x = 2

my problem is i understand how to use the disk method when u flip on the x or y axis.. but when u do it on x = 1 or x =2 for the flip, i dont know how to understand and approach it!

2. first note that if y = 2x, then x = y/2 .

hope you have made a sketch and know that this problem involves horizontal disks with respect to y.

for rotation about x = 1, your radius of revolution is a horizontal segment from the line x = 1 to the line x = y/2.

since x = 1 is to the right of x = y/2, the length of the radius is 1 - y/2.

so, your volume found by disks is ...

$V = \pi \int_0^2 \left(1 - \frac{y}{2}\right)^2 \, dy$

for rotation about x = 2, note that this will involve the washer method w/r to y.

your inner radius goes from x = 2 to x = 1 ... r = 2-1 = 1

your outer radius goes from x = 2 to x = y/2 ... R = 2 - y/2
$
V = \pi \int_0^2 \left(2 - \frac{y}{2}\right)^2 - (1)^2 \, dy$

3. Originally Posted by skeeter
first note that if y = 2x, then x = y/2 .

hope you have made a sketch and know that this problem involves horizontal disks with respect to y.

for rotation about x = 1, your radius of revolution is a horizontal segment from the line x = 1 to the line x = y/2.

since x = 1 is to the right of x = y/2, the length of the radius is 1 - y/2.

so, your volume found by disks is ...

$V = \pi \int_0^2 \left(1 - \frac{y}{2}\right)^2 \, dy$

for rotation about x = 2, note that this will involve the washer method w/r to y.

your inner radius goes from x = 2 to x = 1 ... r = 2-1 = 1

your outer radius goes from x = 2 to x = y/2 ... R = 2 - y/2
$
V = \pi \int_0^2 \left(2 - \frac{y}{2}\right)^2 - (1)^2 \, dy$
hm ok i understand part a.. but can u help clarify how for part b u got a 2 and also a -1 at the end thanks!

4. ok i m guessing the two comes from the fact your flippen it over the 2 which is "on top" with respect to y.. but the -1 ^2 i m not seeing where that is coming from

5. the integral method for horizontal washers is

$V = \pi \int_c^d (R^2 - r^2) \, dy$

I told you how I determined R and r in my previous post.

6. Originally Posted by skeeter
the integral method for horizontal washers is

$V = \pi \int_c^d (R^2 - r^2) \, dy$

I told you how I determined R and r in my previous post.

i know but it is makin no sense to me , hm is there another way to explain how to set this problem up?