Find a formula for the area A(x) of the cross sections of the the solid that are

Question

Find a formula for the area A(x) of the cross sections of the the solid that are perpendicular to the x-axis. The solid lies between planes perpendicular to the x-axis perpendicular at x = 0 and x = 4. The cross sections to the perpendicular to the x-axis between these planes run from y = squareroot x to squareroot x. (a) The cross sections are circular disks with diameters in the xy-plane. A(x) = . (b) The cross sections are squares with bases in the xy-plane. A(x) = . (c) The cross sections are squares with diagonals in the xy-plane A(x) = . (d) The cross sections are lateral triangles with bases in themiddotxy-plane. A(x) =

Explanation / Answer

Solution : (A)

Diameter = x - (-x) = 2x

Radius = 2x / 2 = x

Area = r2 = (x)2 = x

From x = 0 to x = 4

=> 04 x dx = [x2/2]04 = (/2) (42 - 0) = 8

Solution : (B)

Side = x - (-x) = 2x

Area = (2x)2 = 4x

From x = 0 to x = 4

=> 04 4x dx = [4x2/2]04 = (2) (42 - 0) = 32

Solution : (C)

Diagonal ; => {(x)2 + (-x)2} = (2x)

Area = [(2x)]2 = 2x

From x = 0 to x = 4

=> 04 2x dx = [2x2/2]04 = (42 - 0) = 16

Solution : (D)

Side = x - (-x) = 2x

Area = (3 / 4) (2x)2 = 3 * x

From x = 0 to x = 4

=> 04 3 * x dx = [3 * x2/2]04 = (3 / 2) (42 - 0) = 83

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