Help Pls: Volume Using Integrals

[apoptosis]

Hello! I have an integral problem here dealing volume. I think I have a good idea on how to get to the answer, but I'm stuck on finding the antiderivative. Here's my work! Thanx in advance to any help!

Homework Statement

There is a solid lying between planes perpendicular to the x-axis from x= -$$\pi$$/3 to $$\pi$$/3. Cross sections on the x-axis are perpendicular to the x-axis are circular disks where the diameter goes from the curve y=tanx to y=secx. Find the volume (by slicing).

Homework Equations

All righty. So far I have graphed the two curves from the two x endpoints.
since y=secx is above y=tanx
L=f(x)=secx-tanx

A(x)=($$\pi$$D^2)/4
Therefore A(x)=($$\pi$$(secx-tanx)^2)/4

For volume:
I have the integral from $$\pi$$/3 to -$$\pi$$/3 A(x)

The Attempt at a Solution

For the solutionn, I know that I have to find the antiderivative of A(x) and take the difference from the two endpoints. Which the the step that I'm stuck at if I have done everything else correct.

Could I have missed a step, or actually am off track with this problem?
Thanks!

 

[Dick]

I think you're a little off track. You want the difference of the areas of the two disks, right? I would make that related to sec^2(x)-tan^2(x) rather than (sec(x)-tan(x))^2. There's an interesting relation between those two trig function that will make your life a lot easier.

 

[rohanprabhu]

I didn't really get ur question.. especially what you meant by: "Cross sections on the x-axis are perpendicular to the x-axis are circular disks where the diameter goes from the curve y=tanx to y=secx". Sorry for that..

but if you want this:

$$I = \int \frac{\pi}{4}~(sec(x) - tan(x))^2dx$$

then, i can help:

$$I = \int \frac{\pi}{4}~(sec^2x + tan^2x - 2sec(x)tan(x))dx$$

Also,

$$tan^2x = sec^2x - 1$$

Hence,

$$I = \int \frac{\pi}{4}~(2sec^2x - 1 - 2sec(x)tan(x))dx$$

I substituted, $sec^2x$ for $tan^2x$ than the other way round because $sec^2x$ is the antiderivative of tanx. Hence, the integral becomes easier.

$$I = \frac{\pi}{4} \times (2tan(x) - x - 2sec(x))$$

now you can put in the limits and solve.

 

[rohanprabhu]

ok.. i re-read the question and now I'm getting what the question meant. I tried getting the equation for volume and it is the same as what you've got.. and well.. as I've already integrated it... put the values and see if it gets you the correct answer.

I'm getting the answer as $\frac{2\pi}{3}$

EDIT:

I tried to visualize this in 3D and this is what I've got:

http://img518.imageshack.us/img518/5234/solidxzf2.jpg

but the problem is that the cross sections are elliptical. Is it because of distorted plotting? Because i don't really see any parameter which might control the eccentricity of the cross section.

I have provided the function used to plot this curve in the image. It gives the z-coordinate as a

function of x and y.

 

[apoptosis]

Thanks rohan for the posts.
Sorry about the wordiness of the problem, but the question i have here on paper is even more convoluted.
I now understand how the trig identities led to simplifying the integrals which allows for the antiderivative. The diagrams also helped me visualize the object much better.

However, I've taken that and found the difference from $$\pi$$/3 to -$$\pi$$/3 and got a very different answer.
Here's my work so far on it: perhaps I've done something wrong
$$\pi$$/4(2tanx-x-2secx)
$$\pi$$/4[(2$$\sqrt{3}$$-$$\pi$$/3-4)-(-2$$\sqrt{3}$$+$$\pi$$/3-4)]
$$\pi$$/4(4$$\sqrt{3}$$-2$$\pi$$/3)

 

[rohanprabhu]

perhaps I've done something wrong
$$\pi$$/4(2tanx-x-2secx)
$$\pi$$/4[(2$$\sqrt{3}$$-$$\pi$$/3-4)-(-2$$\sqrt{3}$$+$$\pi$$/3-4)]
$$\pi$$/4(4$$\sqrt{3}$$-2$$\pi$$/3)

Actually.. nope... i made a blunder. I used a cas to solve the function.. but i didn't input the tan(x) function in it.. your answer is correct.. :D

 

[Dick]

Apparently, I don't understand either one of you. Isn't it just 2*pi/3?

 

[rohanprabhu]

Apparently, I don't understand either one of you. Isn't it just 2*pi/3?

$\frac{2\pi}{3}$ is only for the inner integral. I forgot to multiply it by the $\frac{\pi}{4}$ that is outside the integral.

this image shall help you:

http://img503.imageshack.us/img503/7828/function1qw6.jpg