Explanation Given: p ( r ) = 150 ( r 2 + 2 ) − 2 deer per square kilometer 2 ≤ r ≤ 5 Calculation: We have been given that, 2 ≤ r ≤ 5 and radial density = deer per square kilometer To calculate the population in the given circle, thin rings inside the circle are considered, as shown below, Now, if we straightened out each ring we get rectangular strips, with length as the circumference of that ring 2πr (r is radius of the ring) and breadth as the difference in radius dr Therefore, area of ring = 2πr*dr square km Therefore, Population of ring = area of ring* radial population density Therefore, total population, P in thousands = sum of (area of ring*radial population density) = integral of 2πrdr* radial population density p(r) P = ∫ 2 5 2 π r p ( r ) d r P = 2 π ∫ 2 5 r p ( r ) d r P = 2 π ∫ 2 5 r 150 ( r 2 + 2 ) − 2 d r L e t , u = r 2 + 2 d u = 2 r d r r d r = d u 2 P = 300 π ∫ 2 5 u − 2 d u 2 P = 300 π × 1 2 ∫ 2 5 u −

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 6.2, Problem 33E

To determine

To calculate: the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is $p (r) = 150 {(r^{2} + 2)}^{- 2}$ deer per square kilometer

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Chapter 6.2, Problem 32E

Chapter 6.2, Problem 34E

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Chapter 6 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 6.1 - Prob. 1PQ Ch. 6.1 - Prob. 2PQ Ch. 6.1 - Prob. 3PQ Ch. 6.1 - Prob. 4PQ Ch. 6.1 - Prob. 5PQ Ch. 6.1 - Prob. 6PQ Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E

Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E Ch. 6.1 - Prob. 11E Ch. 6.1 - Prob. 12E Ch. 6.1 - Prob. 13E Ch. 6.1 - Prob. 14E Ch. 6.1 - Prob. 15E Ch. 6.1 - Prob. 16E Ch. 6.1 - Prob. 17E Ch. 6.1 - Prob. 18E Ch. 6.1 - Prob. 19E Ch. 6.1 - Prob. 20E Ch. 6.1 - Prob. 21E Ch. 6.1 - Prob. 22E Ch. 6.1 - Prob. 23E Ch. 6.1 - Prob. 24E Ch. 6.1 - Prob. 25E Ch. 6.1 - Prob. 26E Ch. 6.1 - Prob. 27E Ch. 6.1 - Prob. 28E Ch. 6.1 - Prob. 29E Ch. 6.1 - Prob. 30E Ch. 6.1 - Prob. 31E Ch. 6.1 - Prob. 32E Ch. 6.1 - Prob. 33E Ch. 6.1 - Prob. 34E Ch. 6.1 - Prob. 35E Ch. 6.1 - Prob. 36E Ch. 6.1 - Prob. 37E Ch. 6.1 - Prob. 38E Ch. 6.1 - Prob. 39E Ch. 6.1 - Prob. 40E Ch. 6.1 - Prob. 41E Ch. 6.1 - Prob. 42E Ch. 6.1 - Prob. 43E Ch. 6.1 - Prob. 44E Ch. 6.1 - Prob. 45E Ch. 6.1 - Prob. 46E Ch. 6.1 - Prob. 47E Ch. 6.1 - Prob. 48E Ch. 6.1 - Prob. 49E Ch. 6.1 - Prob. 50E Ch. 6.1 - Prob. 51E Ch. 6.1 - Prob. 52E Ch. 6.1 - Prob. 53E Ch. 6.1 - Prob. 54E Ch. 6.1 - 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Prob. 39CRE Ch. 6 - Prob. 40CRE Ch. 6 - Prob. 41CRE Ch. 6 - Prob. 42CRE Ch. 6 - Prob. 43CRE Ch. 6 - Prob. 44CRE Ch. 6 - Prob. 45CRE Ch. 6 - Prob. 46CRE Ch. 6 - Prob. 47CRE Ch. 6 - Prob. 48CRE Ch. 6 - Prob. 49CRE Ch. 6 - Prob. 50CRE Ch. 6 - Prob. 51CRE Ch. 6 - Prob. 52CRE Ch. 6 - Prob. 53CRE Ch. 6 - Prob. 54CRE

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the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is p ( r ) = 150 ( r 2 + 2 ) − 2 deer per square kilometer

Solution Summary: The author explains that if the radial density is p ( r ) = 150 2 deer per square kilometer, the population in the given circle is calculated.

Videos

To calculate: the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is $p (r) = 150 {(r^{2} + 2)}^{- 2}$ deer per square kilometer

Want to see the full answer?

Chapter 6 Solutions

the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is p ( r ) = 150 ( r 2 + 2 ) − 2 deer per square kilometer

Solution Summary: The author explains that if the radial density is p ( r ) = 150 2 deer per square kilometer, the population in the given circle is calculated.

Videos

To calculate: the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is p(r)=150(r2+2)−2 deer per square kilometer

Want to see the full answer?

Chapter 6 Solutions

To calculate: the number of deer within a 2 km to 5km radial distance of the clearing if the radial density is $p (r) = 150 {(r^{2} + 2)}^{- 2}$ deer per square kilometer