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Unformatted text preview: Math 21BB  Homework Set 5 Section 6.1: 4. Let’s consider the area of a cross section of the solid when 1 ≤ x ≤ 1. The diameter of the circular cross section at x is given by Diameter = ( 2 x 2 ) x 2 = 2 2 x 2 . Thus, the area of the cross section is given by Area = π · 1 2 ( 2 2 x 2 ) 2 = π · ( 1 x 2 ) 2 = π · ( 1 2 x 2 + x 4 ) . Volume = Z 1 1 π · ( 1 2 x 2 + x 4 ) dx = π Z 1 1 ( 1 2 x 2 + x 4 ) dx = π · x 2 3 x 3 + 1 5 x 5 1 1 = π · 1 2 3 + 1 5 π · 1 + 2 3 1 5 = π · 2 4 3 + 2 5 = 16 15 π 7. a. Let’s consider the area of a cross section of the solid when 0 ≤ x ≤ π . The base of the triangular cross section at x is given by Base = 2 √ sin x . The height of the triangle is given by Height = √ 3 · √ sin x . ( Use properties of equilateral triangles ) Therefore, the area of the cross section is given by Area = ( 1 2 ) 2 √ sin x √ 3 √ sin x = √ 3sin x . 1 Volume = Z π √ 3sin xdx = √ 3cos x π = √ 3 ( √ 3) = 2 √ 3 b. Let’s consider the area of a cross section of the solid when 0 ≤ x ≤ π . The base of the square cross section at x is given by Base = 2 √ sin x . Therefore, the area of the cross section is given by Area = (2 √ sin x ) 2 = 4sin x . Volume = Z π 4sin xdx = 4cos x  π = 4 ( 4) = 8 12. a. Diameter = 2 4 √ 1 x 2 Area = π · h 1 2 2 4 √ 1 x 2 i 2 = π √ 1 x 2 Volume = Z √ 2 / 2 √ 2 / 2 π √ 1 x 2 dx = π Z √ 2 / 2 √ 2 / 2 1 √ 1 x 2 dx = ( π · sin 1 x ) √ 2 / 2 √ 2 / 2 = π · π 4 π ·  π 4 = 2 · π 2 4 = π 2 2 2 b. Diagonal = 2 4 √ 1 x 2 Area = r 1 4 √ 1 x 2 2 + 1 4 √ 1 x 2 2 ! 2 = 2 √ 1 x 2 Volume = Z √ 2 / 2 √ 2 / 2 2 1 x 2 dx = ( 2sin 1 x ) √ 2 / 2 √ 2 / 2 = 2 · π 4 2 ·  π 4 = 4 · π 4 = π 15. We are rotating about the xaxis, so we want to take the integral with respect to x . We want 0 ≤ x ≤ 2. The radius of a cross section is given by Radius = 1 2 x + 1 ( Between graph and xaxis ) Therefore, the area of the cross section is given by Area = π · ( 1 2 x + 1 ) 2 = π · ( 1 4 x 2 x + 1 ) Volume = Z 2 π · 1 4 x 2 x + 1 dx = π · 1 12 x 3 1 2 x 2 + x 2 = π · 8 12 2 + 2 = 2 π 3 16. We are rotating about the yaxis, so we want to take the integral with respect to y . We want 0 ≤ y ≤ 2. The radius of the cross section is given by Radius = 3 y 2 ( Between the graph and the yaxis ) Therefore, the area of the cross section is given by 3 Area = π · ( 3 y 2 ) 2 = π · 9 y 2 4 Volume = Z 2 π · 9 y 2 4 dy = π · 3 y 3 4 2 = 6 π 20. We are rotating about the xaxis, so we want to take the integral with respect to y . We want 0 ≤ x ≤ 2. The radius of the cross section is given by Radius = x 3 Therefore, the area of the cross section is given by Area = π ( x 3 ) 2 = π · x 6 Volume = Z 2 π · x 6 dx = π 7 x 7 2 = 128 π 7 29. We are rotating about the line y = √ 2, so we want to take the integral with respect to x . We want 0 ≤ x ≤ √ 2. The radius of the cross section2....
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This note was uploaded on 11/17/2010 for the course MAT 21B MAT 21B taught by Professor Mat21b during the Spring '07 term at UC Davis.
 Spring '07
 MAT21B

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