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Thomas Edison State University MAT-232-OL009 Written Assignment 1. Section 5.2 2) Find the volume of the solid with cross-sectional area A(x). A(x)= 10 e 0.01 x , 0 ≤x ≤ 10 V= a b A ( x ) dx V= 0 10 10 e 0.01 x dx , u=0.01x, du=0.01 dx V=10 0 10 e u 100 du Since e u du = e u , = 10 100 e u = 1000 e u = 1000 e x 100 For the boundaries of 0 ≤x ≤ 10 V=1000 e 10 100 1000 e 0 100 = 1000 e 10 100 = 1000 ( e 1 10 1 ) 6) Find the volume of a pyramid of height 160 feet that has a square base of side 300 feet. These dimensions are half those of the pyramid in example 2.1. How does the volume compare? 10) A dome “twice as big” as that of exercise 9 has outline y=120- x 2 120 for -120 ≤x ≤ 120. (units of feet. Find its volume.
V= 120 x 2 120 = 2 π ( 60 ( x 2 ) x 4 480 ) = 2 π ( 60 120 2 120 4 480 ) = 864000 π ft 3 ¿ x ¿ xy dx = 2 π 0 120 ¿ 2 π a b ¿ 12) A pottery jar has circular cross sections of radius 4-sin x 2 inches for 0 ≤x ≤ 2 π . Sketch a picture of the jar and compute its volume. V= 16 8sin x 2 + sin 2 x 2 π 0 2 π ( 4 sin x 2 ) dx = π 0 2 π ¿ dx = π 0 2 π ( 16 8sin x 2 + 1 2 cos x 2 ) dx = ¿ π ( 16 x + 16cos x 2 + x 2 sin x 2 ) = π ( 32 π 16 + π 16 ) = 32 π 2 32 π 18) Compute the volume of the solid formed by revolving the region bounded by y= x 2 , y = 4 x 2 about (a) the x-axis; (b) y-4. x 2 = 4 x 2 ; 2 x 2 = 4 ; x 2 = 2, so x = ± 2 , y = 4,0 a) Inner-radius= x 2 , outer-radius = 4 x 2 Because our boundaries are ± 2 ,wecanmultiply theintegralby 2, have theboundariesbe 0 ≤ x≤ 2 V=2 4 x 2 ( ¿ ) ¿ ¿ 2 x 4 ¿ π ¿ 0 2 ¿ b) inner-radius=4-4+ x 2 = x 2 outer-radius=4- x 2 again, because our boundaries are ± 2 , we will just multiply our integral by 2 and have the boundaries be 0 ≤x ≤ 2
v= π ( ( 4 x 2 ) 2 x 4 ) dx = ¿ 2 π ( 16 x 8 x 3 3 ) = 2 π ( 16 2 8 2 3 3 )= 64 2 3 π 2 0 2 ¿ 20) Compute the volume of the solid formed by revolving the region bounded by y= x 2 x = y 2 about (a) the y-axis; (b) x=1 The intersection point: { y = x 2 x = y 2 : x- x 4 = 0 ; x ( 1 x 3 ) = x ( x 1 ) ( x 2 + x + 1 ) = 0 ; x 1 = 0, x 2 = 1 a) y= x 2 ; x = ± y Inner-radius= y 2 ,outer radius = y V= π 0 1 ( y y 4 ) dy = π ( y 2 2 y 5 5 ) = π ( 1 2 1 5 ) = 3 5 π b) Inner-radius=1- y , outer-radius=1- y 2 V= π 0 1 ( ( 1 y 2 ) −( 1 2 ) 2 ) dy = π 0 1 ( ( 1 2 y 2 + y 4 1 + 2 y y ) ) V= π ( 2 y 3 3 + y 5 5 + 4 y 3 2 3 + y 2 2 ) = 11 30 π 26) Let R be the region bounded by y= x 2 and y=4. Compute the volume of the solid formed by revolving R about the given line.

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