# Math Help - Volume of the solid revolving about y-axis

1. ## Volume of the solid revolving about y-axis

This involves trig functions so I am not sure if I am on the right track

y = (tan^2 x)/x 0 <= x <= pi/4

a = 0; b=pi/4 on the x-axis

so I think I should use the formula 2pi (integral sign "S") x f(x) dx

v = 2 pi S a=0; b=pi/4 x (tan^2 x)/x dx

so I am thinking the x's cancel each other out and I have
v= 2 pi S tan^2 x dx

v = 2 pi ((1/3)sec^6)

Am I on the right track??

Thanks

Calculus Beginner

2. Originally Posted by calcbeg
This involves trig functions so I am not sure if I am on the right track

y = (tan^2 x)/x 0 <= x <= pi/4

a = 0; b=pi/4 on the x-axis

so I think I should use the formula 2pi (integral sign "S") x f(x) dx

v = 2 pi S a=0; b=pi/4 x (tan^2 x)/x dx

so I am thinking the x's cancel each other out and I have
v= 2 pi S tan^2 x dx

v = 2 pi ((1/3)sec^6)

Am I on the right track??

Thanks

Calculus Beginner
You are except for the last part.

$
V = 2 \pi \int_0^{\pi/4}\tan ^2 x\, dx = 2 \pi \int_0^{\pi/4}\left(\sec ^2 x - 1\right)\, dx
$

3. ## next step

Okay so 2pi S a=0; b= pi/4 sec^2 x - 1 dx

= 2pi (tan x - x) pi/4 and 0

2pi (1 - (pi/4) - 0 - 0) = 2pi - (1/2)pi^2

Is this correct?

Thanks

Calculus beginner

4. Originally Posted by calcbeg
Okay so 2pi S a=0; b= pi/4 sec^2 x - 1 dx

= 2pi (tan x - x) pi/4 and 0

2pi (1 - (pi/4) - 0 - 0) = 2pi - (1/2)pi^2

Is this correct?

Thanks

Calculus beginner
Looks good to me