Triple Integration of a Strange Cylinder

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SUMMARY

The discussion focuses on calculating the volume of the region defined by the inequalities x² + y² ≤ a² and 0 ≤ z ≤ x using cylindrical coordinates. The user initially integrates with bounds from 0 to a, 0 to π, and 0 to rcos(θ), resulting in a volume of zero due to symmetry about the x-axis. The correct approach involves integrating from θ = -π/2 to π/2 to account for the volume above the z = 0 plane, effectively doubling the calculated volume to obtain the correct result.

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  • Familiarity with symmetry in mathematical functions
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Rocker222
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Hey all,

If anyone has some hints on how to do this one it would be much appreciated:

Find the volume of the region given by x2 + y2 ≤ a2, 0 ≤ z ≤ x.

So I've gone to cylindrical polars, and threw in the Jacobian, r. If I integrate with my bounds being:

0 to a, 0 to pi, 0 to rcos(theta) of r dz d(theta) dr I'm getting 0 which can't be. What is it that I'm missing here. Any hint would be great.
 
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So you are integrating
\int_{r=0}^a\int_{\theta= 0}^\pi\int_{z= 0}^{r cos(\theta)} r dzd\theta dr[/itex]?<br /> <br /> If you look closely at this situation you will see that this is symmetric about the x-axis and so, with z= x as a boundary, half the volume is below the z= 0 plane, half above. Since your integral treats the volume below the z= 0 plane as negative, those cancel. The true &quot;volume&quot; is twice the volume above the z-axis. You get the volume above the z= 0 plane by assuring that x is positive: by taking \theta from -\pi/2 to \pi/2.
 
Thank you very much. That's just the bit I needed. Sorry about not using the Latex code. I'm new to this site.
 

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