Determine the volume generated when the ellipse 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?
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.
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