# Homework Help: Calculating volume

1. Jun 1, 2006

### suspenc3

Hi, im having a little trouble with this question, im trying to find the volume of the solid after revolving it.

y=sec x, y=1, x=-1, x=1, about the x-axis

what I did:

$$A(x)= \frac{ \pi}{(cosx)^2}$$

and then I integrated from -1 to 1...(my integral came out to $$\frac{-1}{sinx}$$)
and I got an answer that came out negative..

can anyone help?

2. Jun 1, 2006

### TD

Have you made a sketch of the problem?

Don't forget that you're only considering the area between y = sec(x) and y = 1!

3. Jun 1, 2006

### suspenc3

ohhhhhh..yeah i think i read the question wrong..il try again

4. Jun 1, 2006

### TD

Ok, a sketch will really help. Also, (tan(x))' = sec²(x), in case you had forgotten

5. Jun 1, 2006

### suspenc3

ok..i tried it again..but im still getting it wrong..

heres what i have:

$$A(x)=\pi(1-(secx)^2)$$

and then i integrated from -1 to 1 and then i ended up with somethign around 14

6. Jun 1, 2006

### TD

Why 1-sec²x ?

7. Jun 1, 2006

### suspenc3

im not really sure how to do this one..dont you do this..A=pi(outer radius)^2 - pi(inner radius)^2..i think its called the washer method

8. Jun 1, 2006

### TD

Yes, but on [-1,1], which one is the 'outer radius'? I believe sec(x) >= 1 there...

Remember that sec(x) = 1/cos(x) and cos(x) is bounded between [-1,1]. So without looking at the sign, in absolute value, |sec(x)| will never be smaller than 1.

9. Jun 1, 2006

### suspenc3

yeah..that makes sense..i just looked at my sketch..I drew it wrong

10. Jun 1, 2006

### suspenc3

i got something around 3.5...sounds better

11. Jun 1, 2006

### TD

I'm getting the same

$$\pi \int\limits_{ - 1}^1 {\sec ^2 x - 1dx} = 2\pi \left( {\tan 1 - 1} \right) \approx 3.5$$

12. Jun 1, 2006

### suspenc3

ok, thanks, IM having trouble with another one..Im having rouble seting it up. y=x and y=0, x=2, x=4 about x=1

I drew the two lines..its a pretty simple graph but what does it mean to revolve it around x=1?and do i still have to do the outer radius-inner radius?

13. Jun 1, 2006

### TD

The formula for the volume of a solid of revolution, rotated about the x-axis was:

$$\pi \int\limits_a^b {y^2 dx}$$

Here, y = f(x). By using inner/outer radii and substracting, you could use this formula as well when the rotation axis was parallel to the x-axis.
Now, for revolving about the y-axis, or a line parallel to that such as x = 1, the formula logically becomes:

$$\pi \int\limits_a^b {x^2 dy}$$

And here, x = f(y).

In your problem, I think you'll have to split it in two integrals.

14. Jun 1, 2006

### suspenc3

hrmm..still confused..im never split it into two before

15. Jun 1, 2006

### TD

Well, your 'outer radius' will be x = 4 all the time. But the inner radius will change. Between y = 0 and y = 2, the inner radius is x = 2. But from y = 2 till y = 4, the inner radius will be y = x. If you have a good sketch, you should be able to see this