# Find the volume

1. Nov 1, 2010

### kari82

Find the volume of the solid generated by revolving the region about the given line

The region in the first quadrant bounded above by the line y=1, below by the curve y=√(sin6x), and on the left by the y-axis, about the line y=1

a) pi/6 + 6
b)pi/12 - 1/6
c)pi^2/12 - pi/6
d)pi^2/12 + pi/6

I set up two possible integrals

V=pi∫((arcsin y^2)/6)^2 dy from y=0 to y=1

V=pi∫(√ (sin6x))^2 dx from x=0 to x=pi

2. Nov 1, 2010

### micromass

The volume by rotating a function f around the X-AXIS, is

$$\pi\int_0^2{f(x)^2dx}$$

But you're rotating around the axis y=1. You have to slide down the function first. So the function you should be rotating around the X-axis is $$f(x)=\sqrt{\sin(6x)}-1$$.

3. Nov 1, 2010

### kari82

Thanks! Is that the function i need to use? When I put that in my calculator from 0 to 2 still doesnt give me any of the answers.. :-( what am i doing wrong??

4. Nov 1, 2010

### micromass

No, the ranges in my formula are wrong. You don't have to do it from 0 to 2. You need to do it from 0 to when f becomes 0 (sketch the function, thatll make it easier)

5. Nov 1, 2010

### kari82

I was wondering where did you get 2 from.. Thanks! I'll try that..

6. Nov 1, 2010

0 to pi/12?

7. Nov 1, 2010

### micromass

Yes, that seems right.

8. Nov 1, 2010

### kari82

thank you so much!