# Calculus area problem

• Feb 15th 2007, 07:01 PM
turtle
Calculus area problem
Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are
a) circular disks with diameters running from the curve y = tanx to the curve y = sec x
b) squares whose bases run from the curve y = tan x to the curve y = sec x
• Feb 16th 2007, 05:55 AM
CaptainBlack
Quote:

Originally Posted by turtle
Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are
a) circular disks with diameters running from the curve y = tanx to the curve y = sec x

As sec(x)>tan(x) for x in (-pi/3, pi/3), the volume of the solid is:

int pi (sec(x)-tan(x))^2 /4 dx = sqrt(3) pi - pi^2/6 ~=3.7965

RonL
• Feb 16th 2007, 05:58 AM
CaptainBlack
Quote:

Originally Posted by turtle
Can anyone help me with this problem. I'm stuck, and I don't know what else to try. Thanks for any help.

The solid lies between planes perpendicular to the x-axis at x = -pi/3 and x = pi/3. The cross sections perpendicular to the x-axis are

b) squares whose bases run from the curve y = tan x to the curve y = sec x

Same as part (a) except instead of integrating pi (sec(x)-tan(x))^2/4 from
-pi/3 to pi/3 you will integrate (sec(x)-tan(x))^2/4 from -pi/3 to pi/3.

RonL
• Mar 15th 2010, 03:00 PM
pocketasian
This is an old question, but how would you find this analytically for part a?

For the volume, so far I have:
(pi/4) times the integral from -pi/3 to pi/3 of (2(secx)^2 - 2secxtanx - 1)dx

I tried using u-substitution, making u=secx and du=secxtanxdx, but I got stuck. Will someone help me solve this by hand? Thanks.
• Mar 15th 2010, 09:44 PM
CaptainBlack
Quote:

Originally Posted by pocketasian
This is an old question, but how would you find this analytically for part a?

For the volume, so far I have:
(pi/4) times the integral from -pi/3 to pi/3 of (2(secx)^2 - 2secxtanx - 1)dx

I tried using u-substitution, making u=secx and du=secxtanxdx, but I got stuck. Will someone help me solve this by hand? Thanks.

$[\sec(x)]^2$ is a standard integral, it integrates to $\tan(x)+c$. For $\sec(x)\tan(x)$ replace the trig functions by $\sin$ 's and $\cos$ 's and you are left with the derivative of $1/\cos(x)$.

CB