Integral and cylindrical shell

1. May 17, 2006

merced

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

y = $$x^2$$
y = 4
x = 0

0 <= x <= 2

So I drew these:

I don't know how to determine which region to use (above or below
y = x^2 ). Either seems like it will work.

Last edited: May 17, 2006
2. May 17, 2006

Curious3141

Think (as the question advised) in terms of cylindrical shells (elements). You are rotating about the y-axis. What is the radius of the cylinder? What is the height? Hence determine the volume of that element, and set up the integral.

3. May 17, 2006

Geekster

The region above y=x^2 is sort of in the shape of a bullet pointed straight down the y-axis, and the region below y=x^2 forms a sort of dog dish shaped object. Notice that x=2 is not a bound for the curve. Look at what does bound the curve and that is what you need to find the volume of.

4. May 17, 2006

merced

Actually, x = 2 is a bound because 0 <= x <= 2. Yeah, the book gives the area for the top part...but I don't see why it can't be the lower part.

If it was the lower part, you still could use cylindrical shells, right?

5. May 17, 2006

Geekster

I think I see where you are getting confused...there is a difference between what x varies over, and what the bounds are.

In this case x varies from 0 to 2. That is different from a bound. Suppose the curve y=x^2 was bounded by y=0, and x=1, but x varied from 0 to a billion? It doesn't much matter how x varies so long as the problem makes sense. In this case, x=2 is not a bound, in which case you have the curve y=x^2, y=4 and x=0 (y-axis is x=0) as bounds. Draw that region, then rotate it about the y-axis and you have your volume. Set up the integral and evaluate.