1. ## Volume problem

Determine the volume generated when the ellipse $\frac {x^2} {9} +y^2=1$ is rotated around the line y=5

In my solution book, it says

Translating the ellipse -5 units along the y-axis

Why -5? Aren't you supposed to translate it 5 units?

2. ## Translations in General

Well You are translating up 5 units by using (y-5)

Remeber (x-a) translates a graph a units to the right and (x+5) translates to the left 5 units.
similarly (y-a) translates up a units and (y+a) translates down a units.

Think of the parabola y = x^2 with vertex at (0,0)

For y = (x-2)^2 the vertex is at (2,0) where y = 0 . A translation to the right of 2 units.

Hope this helps.

3. Originally Posted by chengbin
Determine the volume generated when the ellipse $\frac {x^2} {9} +y^2=1$ is rotated around the line y=5

In my solution book, it says

Translating the ellipse -5 units along the y-axis

Why -5? Aren't you supposed to translate it 5 units?
Isn't there something after "translating the ellipse -5 units along the y-axis"?! Why translate at all? You don't have to translate it at all. But translating it down 5 shifts the line y= 5 down to y= 0. Instead of translating around y= 5, you will be rotating around the x-axis, which, presumably you have experience in doing that. It is not really necessary to translate at all.

4. ## Volume problem

I think I misunderstood you original question.

by translating down 5 units and rotating about the x-axis you don't seem to simplify the problem.

You have x^2/9 +(y+5)^2 = 1

solving for y the upper half is y = -5+ (1-x^2/9)^1/2 and the lower half is
y = -5- (1-x^2/9)^1/2

You have a washer with outer radius y = -5- (1-x^2/9)^1/2 and inner radius y = -5+ (1-x^2/9)^1/2

Without translating you have a washer with outer radius y = 5+ (1-x^2/9)^1/2 and inner radius 5- (1-x^2/9)^1/2.

What I'd reccomend is the theorem of Pappus V =2(pi)bA
Where A is the Area of the ellipse pi(3)(1) and b is the distance from the
center of mass to the line y = 5. since the center of mass is (0,0)
You have 30 (pi)^2