Integral and cylindrical shell

[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:
http://img131.imageshack.us/img131/9225/math28qi.th.jpg
http://img68.imageshack.us/img68/866/math21ow.th.jpg

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

 

[Curious3141]

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

$$0\leq x\leq 2$$

So I drew these:
http://img131.imageshack.us/img131/9225/math28qi.th.jpg
http://img68.imageshack.us/img68/866/math21ow.th.jpg

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

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.

 

[Geekster]

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

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.

 

[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?

 

[Geekster]

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.

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.