What is the volume obtained by rotating a region bounded by a given curve?

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Homework Help Overview

The discussion revolves around finding the volume obtained by rotating a region bounded by the curves y = sec(x) and y = cos(x) over the interval [0, π/3]. The rotation is specified about the line y = -1.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of the volume integral, with one participant expressing their approach to defining the outer and inner radii. There is also a question regarding the inclusion of the axis of rotation in the problem statement.

Discussion Status

Some participants are clarifying the axis of rotation and its implications on the setup of the integral. There is acknowledgment of a mistake in the original problem statement regarding the axis, and a few participants are exploring the correct interpretation of the radii used in the volume calculation.

Contextual Notes

There is a noted lack of clarity in the original problem statement, particularly concerning the axis of rotation, which has led to some confusion in the setup of the integral. Participants are encouraged to provide complete information when posting problems.

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Homework Statement



Find the volume obtained by rotating the regon boudned by the given curve about the specified axis

Homework Equations





The Attempt at a Solution



y = secx, y = cosx, 0 <= x < = pi/3

This is what I set up.

V = ∏∫ (secx +1 )^2 - (cosx +1)^2 dx

I said R was secx - (-1) and r is cosx -(-1)
 
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Jbreezy said:

Homework Statement



Find the volume obtained by rotating the regon boudned by the given curve about the specified axis

Homework Equations





The Attempt at a Solution



y = secx, y = cosx, 0 <= x < = pi/3

This is what I set up.

V = ∏∫ (secx +1 )^2 - (cosx +1)^2 dx

I said R was secx - (-1) and r is cosx -(-1)

Looks OK, assuming you are rotating the region about the line y = -1, which is info you didn't provide. When you post a problem, remember to include the complete problem statement.
 
Yeah I'm doing it about x = -1. Thanks dude. Sorry.
 
Why the 1's?
The radii in the cross-sectional disks are specified with the function values.
Thus, your integrand is sec^2(x)-cos^2(x)

I assumed it was about y=0
 
Yeah well that is because I made a mistake in my post and left out that is was about y = -1.
 
Jbreezy said:
Yeah I'm doing it about x = -1. Thanks dude. Sorry.

Jbreezy said:
Yeah well that is because I made a mistake in my post and left out that is was about y = -1.

Apparently you aren't clear, either.
 
Ha. Lack of sleep
 

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