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