Triple integral using cylindrical coordinates

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SUMMARY

The discussion focuses on evaluating the triple integral \(\int\int_{Q}\int(x^4+2x^2y^2+y^4)dV\) over a cylindrical solid defined by \(\{(x,y,z)| x^2+y^2 \leq a^2, 0\leq z\leq \frac{1}{\pi}\}\). The conversion to cylindrical coordinates yields the function \(f(r,\theta,z) = r^4\), simplifying the integration process. The correct bounds for the integral are established as \(r\) from 0 to \(a\), \(\theta\) from 0 to \(2\pi\), and \(z\) from 0 to \(\frac{1}{\pi}\).

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  • Cylindrical coordinates in multivariable calculus
  • Understanding of triple integrals
  • Knowledge of integration techniques
  • Familiarity with polar coordinate transformations
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  • Study the application of cylindrical coordinates in triple integrals
  • Learn about the Jacobian in coordinate transformations
  • Explore examples of integrating functions over cylindrical solids
  • Review the properties of integrals in multivariable calculus
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Students and educators in calculus, particularly those focusing on multivariable integration techniques, as well as anyone seeking to understand the application of cylindrical coordinates in solving triple integrals.

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Homework Statement


[tex]\int\int_{Q}\int(x^4+2x^2y^2+y^4)dV[/tex] where Q is the cylindrical solid given by [tex]\{(x,y,x)| x^2+y^2 \leq a^2, 0\leqz\leq\frac{1}{\pi}\}[/tex]

Homework Equations



When I convert to cylindrical I get [tex]f(r,\theta,z) = r^4\cos^2\theta + 2r^4\cos^2\theta\sin^2\theta + r^2\sin^2\theta[/tex], but I just need the bounds for dr, is it? [tex]\int^a_0[/tex]

The Attempt at a Solution


[tex]\int^2\pi_0\int^a_0\int^\frac{1}{\pi}_0 f(r,\theta,z)rdzdrd\theta[/tex]
*the first integral is supposed to be from 0 to 2pi
 
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Your expression for [itex]f(r,\theta,z)[/itex] is incorrect. It might help you to realize that [itex]f(x,y,z)=x^4+2x^2y^2+y^4=(x^2+y^2)^2[/itex]...so [itex]f(r,\theta,z)=?[/itex] :wink:

And yes, [itex]r[/itex] goes from 0 to a.
 
Got it, so [tex]f(r,\theta,z) = r^4[/tex] Much simpler!
 

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