Explanation Given: x 2 + y 2 = 16 , x ≥ 0 , y ≥ 0 , ρ = k ( x 2 + y 2 ) Calculation: Convert the provided equations into polar coordinates by with the substitution as: x = r cos θ y = r sin θ d x d y = r d r d θ The limits of the provided double integral are: 0 ≤ x ≤ 4 0 ≤ y ≤ 16 − x 2 The corresponding limits in polar coordinates will be: 0 ≤ θ ≤ π 2 0 ≤ r ≤ 4 Hence, the provided equations become: r 2 cos 2 θ + r 2 sin 2 θ = 16 ⇒ r 2 = 16 ⇒ r = 4 a n d 0 ≤ θ ≤ π 2 Now, sketch the graph as shown below, Draw the reference axis in polar coordinates. Then, plot the circle of radius 4 units in the first quadrant with equations: r = 4 , 0 ≤ θ ≤ π 2 Further, mark the points on the plot and connect them. Shade the inside area of the graph as the area of interest is the area inscribed by the equations. The final plot is shown below as: Therefore, the mass of the lamina is: m = ∫ 0 4 ∫ 0 16 − x 2 k ( x 2 + y 2 ) d y d x m = ∫ 0 π 2 ∫ 0 4 k r 3 d r d θ m = ∫ 0 π 2 k [ r 4 4 ] 0 4 d θ m = ∫ 0 π 2 64 k d θ Then, m = [ 64 k θ ] 0 π 2 m = 32 π k Now, to find the center of the mass calculate the moment of inertia about both the axes

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ISBN: 9780357092477

Author: Larson

Publisher: Cengage

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Chapter 14.4, Problem 24E

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To Calculate: The mass and center of mass of the lamina bounded by the graphs of the equations x2+y2=16,x≥0,y≥0,ρ=k(x2+y2).

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Chapter 14.4, Problem 23E

Chapter 14.4, Problem 25E

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The mass and center of mass of the lamina bounded by the graphs of the equations x 2 + y 2 = 16 , x ≥ 0 , y ≥ 0 , ρ = k ( x 2 + y 2 ) .

Solution Summary: The author calculates the mass and center of mass of the lamina by converting the equations into polar coordinates.

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To Calculate: The mass and center of mass of the lamina bounded by the graphs of the equations x2+y2=16,x≥0,y≥0,ρ=k(x2+y2).

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Chapter 14 Solutions