# Math Help - Calculus area problem

1. ## 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

2. 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

3. 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

4. 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.

5. 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