Volume of solid by cross-section question?

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SUMMARY

The discussion focuses on calculating the volume of a solid generated by revolving the region bounded by the curves y = csc(x), x = 1/4π, x = 3/4π, and y = 0 around the x-axis. The volume is determined using the formula V = π ∫[csc²(x)] dx from 1/4π to 3/4π. The correct evaluation of the integral leads to V = -π(cot(3π/4) - cot(π/4)), resulting in a positive volume of 2π after resolving the cotangent values.

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  • Familiarity with trigonometric functions, particularly cosecant and cotangent
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  • Ability to sketch regions defined by curves for visual understanding
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Homework Statement



I need to find the region bounded by these curves then find the volume of the solid generated by revolving this region about the x-axis.

y= cscx, x= 1/4pi, x = 3/4pi, y=0

Homework Equations


The Attempt at a Solution



So I managed to sketch this region.. but I have trouble finding the anti-derivative at the end.. so it looks like this:

<br /> <br /> V = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} [(cscx)^2 - (0)^2]dx<br /> <br /> = \pi \left[ -cotx \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}<br /> <br /> = \pi(-cot(\frac{3\pi}{4})-(-cot(\frac{\pi}{4}))<br /> <br /> = -3+1 = -2??<br /> <br />
 
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Your integral and its antiderivative look fine, but what happened to pi? Your problem doesn't seem to be in the integration, but in evaluating cot(x).
Tip: bring the - outside so that you have -pi(cot(x)), evaluated at 3pi/4 and pi/4.

So you have -pi(cot(3pi/4) - cot(pi/4)).
cot(3pi/4 = -1 and cot(pi/4) = 1.

Now what do you get? It should be positive.
 

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