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

In summary, the conversation discusses finding the volume of a solid obtained by rotating a region bounded by specific curves using disks or washers. The attempt at a solution involves using the equation V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx and applying a trig identity. The correct answer is found by taking the antiderivative of the integrand.
  • #1
twoski
181
2

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 [itex]V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx[/itex]

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

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.
 
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  • #2
twoski said:

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 [itex]V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx[/itex]

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

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?
 
  • #3
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.
 
  • #4
twoski said:
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).
 

1. What is the difference between the washer and disk method in finding the volume of a rotating region?

The washer method is used when the cross-section of the solid is a washer shape, while the disk method is used when the cross-section is a disk shape. The washer method is typically used when the region is bounded by two curves, while the disk method is used when the region is bounded by one curve.

2. How do you set up the integral for the washer method?

The integral for the washer method is typically set up as ∫(π(R²-r²))dx, where R is the outer radius of the washer and r is the inner radius of the washer. This integral is then evaluated over the bounds of the region to find the volume.

3. Can the washer and disk method be used to find the volume of any rotating region?

Yes, the washer and disk method can be used to find the volume of any rotating region as long as the region has a washer or disk shape as its cross-section. This method is commonly used in finding the volume of solids of revolution.

4. What if the region has a hole in the center? Can the washer and disk method still be used?

Yes, the washer and disk method can still be used even if the region has a hole in the center. The inner radius (r) in the integral for the washer method would just be equal to the radius of the hole, and the outer radius (R) would be the distance from the center of the hole to the outer curve.

5. Are there any limitations to using the washer and disk method?

The washer and disk method may not be applicable if the region has a cross-section that is not a washer or disk shape. In this case, other methods such as the shell method may be used to find the volume of the rotating region. Additionally, the washer and disk method can only be used for solid objects, not hollow objects.

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