How Many Fragments Are Found Within 10 Kilometers of a Volcanic Eruption?

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SUMMARY

The density of volcanic fragments within a 10-kilometer radius from the eruption center is described by the function D(r) = 1/[sqrt(x) + 2] fragments per square kilometer. To accurately calculate the total number of fragments, a double integral must be performed in polar coordinates, specifically J = ∫(0 to 2π) ∫(0 to 10) (1/(sqrt(x)+2) * x) dx dθ. The correct result of this integration yields approximately 70.424 fragments, correcting the initial miscalculation of 26.294. The integration process involves recognizing the need for a Jacobian factor due to the polar coordinate transformation.

PREREQUISITES
  • Understanding of double integrals in polar coordinates
  • Familiarity with the Jacobian transformation in calculus
  • Knowledge of integrating functions over specified limits
  • Basic concepts of density functions in mathematical modeling
NEXT STEPS
  • Study the application of double integrals in polar coordinates
  • Learn about the Jacobian determinant and its role in coordinate transformations
  • Explore density functions and their integration in real-world scenarios
  • Practice solving problems involving integration of functions over areas
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Mathematicians, physics students, environmental scientists, and anyone interested in the mathematical modeling of volcanic activity and fragment distribution.

radeksrat
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Not really a homework question, but a problem I don't get nonetheless.



The density of fragments lying x kilometers from the center of a volcanic eruption is given by:

D(r) = 1/[sqrt(x) +2] fragments per square kilometer. To 3 decimal places, how many fragments will be found within 10 kilometers of the eruption's center?



I thought I was supposed to integrate the function from 0 to 100*pi, and in doing so I got 26.294, (I got 2[sqrt(x) - 2*ln(sqrt(x)+2)] when i integrated the function) but the answer was given to me as 70.424. The answer could very well be wrong, but I don't know that it is. What, if anything, am I doing wrong here?
 
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in cylindrical coordinates the integral of the density gives the distribution.
In this case the problem requires integrating over an area thus we have a double integral. In polar form J = int(int(f(r,theta)*r*dr)*dtheta) With the appropriate limits. Then J = int(int(1/(sqrt(x)+2),x) from 0 to 10,theta from 0 to 2pi) The result is simpler because theta does not appear inside the integral. The result is 2*pi*int(1/(sqrt(x)+2),x) from 0 to 10. That should be your answer
 
I'm not sure what you mean by ,x) abercrombie, but to be clear, the integral is
\int_{0}^{2 \pi}\int_{0}^{10} \frac{x}{\sqrt{x}+2} dx d\theta
because of the jacobian x.
 

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