Explanation Given: y = e x , y = 0 , x = 0 , x = 1 , and ρ = k Formula used: The mass is given by M a s s , m = ∫ R ∫ d e n s i t y ( x , y ) d A = ∫ R ∫ ρ ( x , y ) d A And the center of mass coordinates are given by ( x ¯ , y ¯ ) = ( M y m , M x m ) where M x = ∫ R ∫ ( y ) ρ ( x , y ) d A and M y = ∫ R ∫ ( x ) ρ ( x , y ) d A Calculation: First, draw the graph for the given equations by performing the following steps: Step 1: Draw the reference axis in xy coordinates. Step 2: Plot the given equations: y = e x , y = 0 , x = 0 , x = 1. Step 3: Mark the points on the plot and connect the points. Step 4: The area of interest is the area inscribed by the equations, so shade the inside area of the graph. Step 5: The final plot is as shown below: The mass of the lamina is: m = ∫ 0 1 ∫ 0 e x k d y d x = ∫ 0 1 [ k y ] 0 e x d x = ∫ 0 1 k e x d x = [ k e x ] 0 1 m = k ( e − 1 ) To find the center of the mass, we have to find the moment of inertia about both axes

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ISBN: 9781337275347

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.4, Problem 17E

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To calculate: The mass and center of mass of the lamina bounded by the graphs of the equations y=ex, y=0, x=0, x=1 for the given density ρ=k.

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Chapter 14.4, Problem 16E

Chapter 14.4, Problem 18E

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The mass and center of mass of the lamina bounded by the graphs of the equations y = e x , y = 0 , x = 0 , x = 1 for the given density ρ = k .

Solution Summary: The author explains the formula used to calculate the mass and center of mass of the lamina.

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To calculate: The mass and center of mass of the lamina bounded by the graphs of the equations y=ex, y=0, x=0, x=1 for the given density ρ=k.

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Chapter 14 Solutions