Find the Volume of a Rotating Region: Washer & Disk Method

[twoski]

Homework Statement

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = sec(x), y = 1, x = 1, x = -1 on the x-axis.

The Attempt at a Solution

This should be ridiculously easy but apparently my answer is wrong?!

To calculate the volume i use the equation $V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx$

From here it should be trivial since this evaluates to $\pi [ tan^{2} x | _{-1}^{1} ] = 0$

But my answer is apparently wrong... Should i be doing something differently? I've followed the formula exactly how it should be followed to my knowledge.

 

[LCKurtz]

Homework Statement

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = sec(x), y = 1, x = 1, x = -1 on the x-axis.

The Attempt at a Solution

This should be ridiculously easy but apparently my answer is wrong?!

To calculate the volume i use the equation $V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx$

From here it should be trivial since this evaluates to $\pi [ tan^{2} x | _{-1}^{1} ]$

But my answer is apparently wrong... Should i be doing something differently? I've followed the formula exactly how it should be followed to my knowledge.

Is the antiderivative of $\sec^2x$ correct? And did you forget the antiderivative of the $-1$? Or did you use a trig identity and forget to integrate it?

 

[twoski]

Whoops. So i apply a trig identity... sec^2(x)-1 = tan^2(x)

Then i antiderive tan^2(x) which gives me... tan(x) - x

So, computing (tan(1) - 1) - (tan(-1) + 1) results in... pi(2tan(1)-2) or 3.5022... And that's right! Thanks for the help.

 

[LCKurtz]

Whoops. So i apply a trig identity... sec^2(x)-1 = tan^2(x)

Then i antiderive tan^2(x) which gives me... tan(x) - x

So, computing (tan(1) - 1) - (tan(-1) + 1) results in... pi(2tan(1)-2) or 3.5022... And that's right! Thanks for the help.

Notice you could have written the answer tan(x)-x directly by taking the antiderivative of your integrand as it stands since the derivative of tan(x) is sec^2(x).