Calculating Volume of Solid Using Triple Integral

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Homework Help Overview

The problem involves calculating the volume of a solid defined by a cylinder and two planes, specifically using a triple integral. The solid is enclosed by the cylinder described by the equation x² + z² = 4 and the planes y = -1 and y + z = 4.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply a cylindrical coordinate system but questions the validity of their approach due to the orientation of the cylinder. Some participants discuss the flexibility of naming coordinate axes and suggest that the mathematical validity remains unchanged regardless of the labels used.

Discussion Status

The discussion is active, with participants exploring different interpretations of the coordinate system and its implications for the volume calculation. Some guidance has been provided regarding the use of cylindrical coordinates and the nature of the volume element, but no consensus has been reached on the best approach.

Contextual Notes

There is a mention of the integration limits being dependent on the geometry defined by the planes, indicating a potential complexity in setting up the triple integral. The discussion also reflects varying levels of familiarity with the concepts involved.

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1. Use a triple integral to find the volume of the given solid.

The solid enclosed by the cylinder x^2 + z^2 = 4 and the planes y = -1 and y + z = 4


This looked like a cylindrical coordinate system to me, except for the fact that it is not cylindrical around the z-axis but the y-axis. I tried to fix this problem by "rotating" my coordinate axes so that my old z-axis would be my new x-axis, x would become y, and y would become z. I'm not sure if this is a valid approach or not.
 
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It makes no difference mathematically how we name the coordinate axes. The coordinate system ##(r,\theta,y)##, with
##x=rcos\theta##
##z=rsin\theta##,
is as valid a cylindrical coordinate system as the usual ##(r,\theta,z)##.

Now the volume element is ##dxdydz=rdrd\theta dy##. Note that because of the plane ##y+z=4## all of the integration limits in the volume calculation are not constants, one of them is a function of ##r## and ##\theta##.
 
Hilbert1 is absolutely correct. However, if you lack his sophistication, you can just rewrite the problem switching the y and z. They are just symbols, right?
 
Thank you both so much!
 

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