Josephood7

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(32*((-5+sqrt(57))/2)^2-41/4((-5+sqrt(57))/2)^4+1/6((-5+sqrt(57))/2)^6)/2

(64/3*((-5+sqrt(57))/2)^3-41/5((-5+sqrt(57))/2)^5+1/7((-5+sqrt(57))/2)^7)/2

(256((-5+sqrt(57))/2)^2-48((-5+sqrt(57))/2)^4-25((-5+sqrt(57))/2)^5+4((-5+sqrt(57))/2)^6-1/8((-5+sqrt(57))/2)^8)/3

((64/3*((-5+sqrt(57))/2)^3-41/5((-5+sqrt(57))/2)^5+1/7((-5+sqrt(57))/2)^7)/2)/((32*((-5+sqrt(57))/2)^2-41/4((-5+sqrt(57))/2)^4+1/6((-5+sqrt(57))/2)^6)/2)

((256((-5+sqrt(57))/2)^2-48((-5+sqrt(57))/2)^4-25((-5+sqrt(57))/2)^5+4((-5+sqrt(57))/2)^6-1/8((-5+sqrt(57))/2)^8)/3)/((32*((-5+sqrt(57))/2)^2-41/4((-5+sqrt(57))/2)^4+1/6((-5+sqrt(57))/2)^6)/2)

 Match the following integrals with the verbal descriptions of the solids whose volumes they give. Put the letter of the verbal description to the left of the corresponding integral.

The region R is bounded by the curves y=5xy=5x, y=8−x2y=8−x2, and the y-axis, and its mass density is δ(x,y)=xyδ(x,y)=xy. To find the center of gravity of the region you would compute ∫∫Rδ(x,y)dA=∫dc∫q(x)p(x)δ(x,y)dydx,∫dc∫q(x)p(x)xδ(x,y)dydx,∫∫Rδ(x,y)dA=∫cd∫p(x)q(x)δ(x,y)dydx,∫cd∫p(x)q(x)xδ(x,y)dydx, and ∫dc∫q(x)p(x)yδ(x,y)dydx∫cd∫p(x)q(x)yδ(x,y)dydx where

Find the centroid (x¯,y¯)(x¯,y¯) of the region bounded by:

 A disk of radius 6 cm has density 10 g/cm22 at its center, density 0 at its edge, and its density is a linear function of the distance from the center. Find the mass of the disk.