Calculating Volume: Solving for y=sec x, -1≤x≤1

[suspenc3]

Hi, I am having a little trouble with this question, I am 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?

 

[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!

 

[suspenc3]

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

 

[TD]

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

 

[suspenc3]

ok..i tried it again..but I am 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

 

[TD]

Why 1-sec²x ?

 

[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

 

[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.

 

[suspenc3]

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

 

[suspenc3]

i got something around 3.5...sounds better

 

[TD]

I'm getting the same

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

A good sketch will lead you to the answer

 

[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?

 

[TD]

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

<br /> \pi \int\limits_a^b {y^2 dx} <br />

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:

<br /> \pi \int\limits_a^b {x^2 dy} <br />

And here, x = f(y).

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

 

[suspenc3]

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

 

[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