# Integral and volume

1. May 16, 2006

### merced

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y = x
y = [tex]\sqrt{x}[\tex]

=

So, I am integrating with respect to x.
Area = [tex]\int^1_{0}\pi[(f(x))^2-(g(x))^2]dx[\tex]

I assumed that f(x) = x and g(x) = [tex]\sqrt{x}[\tex].

However, the book gives f(x) = 1 - x and g(x) = 1 - [tex]\sqrt{x}[\tex].

I don't understand how they got that.

Last edited: May 16, 2006
2. May 16, 2006

### Curious3141

3. May 16, 2006

### Lyuokdea

what axis are you rotating the function around and how does that affect the volumes you will get?

A nicer working problem to go through, which might better illustrate the principals is this: Find the volume generated by rotating the area in between the two functions: y=1 and y=1/2 around the axis y=1. How do the two lines function when they are rotated?

Note: You will be able to check your answer to this problem by comparing the answer with the volume of a cylinder.

Curious: beat me to it I guess on the other thread.

~Lyuokdea