Triple Integral in Cylindrical Coordinates

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SUMMARY

The discussion focuses on evaluating the triple integral \(\int \int \int_E x^2 \, dV\) for the solid defined by the cylinder \(x^2+y^2=1\), above the plane \(z=0\), and below the cone \(z^2=4x^2+4y^2\). The correct bounds for the integral in cylindrical coordinates are established as \(z\) ranging from \(0\) to \(2r\). The user initially attempted the integral with incorrect bounds, specifically using \(-2r\) to \(2r\) for \(z\), which was corrected by another forum member.

PREREQUISITES
  • Cylindrical coordinates in calculus
  • Understanding of triple integrals
  • Knowledge of solid geometry involving cylinders and cones
  • Familiarity with integration techniques in multivariable calculus
NEXT STEPS
  • Study the application of cylindrical coordinates in triple integrals
  • Learn about the geometric interpretation of integrals over solids
  • Practice evaluating integrals with varying bounds in cylindrical coordinates
  • Explore the relationship between different coordinate systems in multivariable calculus
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and integration techniques. This discussion is beneficial for anyone looking to deepen their understanding of triple integrals in cylindrical coordinates.

daveyman
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Homework Statement


Evaluate [tex]\int \int \int_E x^2 \, dV[/tex] where E is the solid that lies within the cylinder [tex]x^2+y^2=1[/tex], above the plane [tex]z=0[/tex], and below the cone [tex]z^2=4x^2+4y^2[/tex].

Homework Equations


In cylindrical coordinates, [tex]x^2+y^2=r^2[/tex] and [tex]x=r\cos{\theta}[/tex].

The Attempt at a Solution


I tried [tex]\int _0^{2\pi }\int _0^1\int _{-2r}^{2r}r^2 cos^2\theta\;\;r\;\;dzdrd\theta[/tex] but I must have messed up the bounds somehow.

Any ideas?
 
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daveyman said:
… above the plane [tex]z=0[/tex]

I tried [tex]\int _0^{2\pi }\int _0^1\int _{-2r}^{2r}r^2 cos^2\theta\;\;r\;\;dzdrd\theta[/tex] but I must have messed up the bounds somehow.

Hi daveyman! :smile:

z goes from zero to 2r. :wink:
 
Of course - it says it right in the problem :-) Thanks!
 

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