Triple Integral Help: Find Mass in Cylindrical Coords

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SUMMARY

The discussion focuses on calculating the mass of a region defined in cylindrical coordinates, specifically where \( r^3 \leq z \leq 1 \) and the density function is given by \( r(r; \theta; z) = 9z \). The integral setup involves evaluating \( \int_0^{2\pi} \int_0^1 \int_{r^3}^1 9z \, dz \, r \, dr \, d\theta \). The conclusion suggests that the correct answer is \( \frac{\pi}{4} \cdot 9^3 \), emphasizing the importance of clarity in problem formulation for accurate computation.

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  • Cylindrical coordinates
  • Triple integrals
  • Density functions in calculus
  • Integration techniques
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  • Review the properties of cylindrical coordinates
  • Study the evaluation of triple integrals
  • Learn about density functions and their applications in mass calculations
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Students and professionals in mathematics, particularly those studying calculus and integral applications, as well as anyone involved in physics or engineering requiring mass calculations in cylindrical coordinates.

bdario1
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Find the mass of the region (in cylindrical coordinates)r^3<=z<=1 , where the density function is r(r;q; z) = 9z.

This is what I got so far

0 <= r<= 1; 0 <= q <= 2p:
hence need to compute the integral
Z from 0 to 2pi Z from 0 to 1 Z r^3 to 1 9zdzrdrdq:
We thus obtain
9p Z 0 to 1(1^2-r^6) r dr


Now this is where I am not able to get to the answer not sure why.
 
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welcome to pf!

hi bdario1! welcome to pf! :smile:

(have an integral: ∫ and a theta: θ and a pi: π and try using the X2 icon just above the Reply box :wink:)
bdario1 said:
Find the mass of the region (in cylindrical coordinates)r^3<=z<=1 , where the density function is r(r;q; z) = 9z.

This is what I got so far

0 <= r<= 1; 0 <= q <= 2p:
hence need to compute the integral
Z from 0 to 2pi Z from 0 to 1 Z r^3 to 1 9zdzrdrdq:
We thus obtain
9p Z 0 to 1(1^2-r^6) r dr


Now this is where I am not able to get to the answer not sure why.

looks ok to me (though not very readable) :confused: …

what is the answer?​
 
I think you should obtain Pi/4 9^3 as answer. Please be more careful when formulating your problem, because it not very readable indeed.
 

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