Solução_Calculo_Stewart_6e

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F. TX.10 CHAPTER 8 PROBLEMS PLUS ¤ 379 7. To find the height of the pyramid, we use similar triangles. The first figure shows a cross-section of the pyramid passing through the top and through two opposite corners of the square base. Now |BD| = b, since it is a radius of the sphere, which has diameter 2b since it is tangent to the oppositesidesofthesquarebase.Also,|AD| = b since 4ADB is isosceles. So the height is |AB| = √ b 2 + b 2 = √ 2 b. We first observe that the shared volume is equal to half the volume of the sphere, minus the sum of the four equal volumes (caps of the sphere) cut off by the triangular faces of the pyramid. See Exercise 6.2.51 for a derivation of the formula for the volume of a cap of a sphere. To use the formula, we need to find the perpendicular distance h of each triangular face from the surface of the sphere. We first find the distance d from the center of the sphere to one of the triangular faces. The third figure shows a cross-section of the pyramid through the top and through the midpoints of opposite sides of the square base. From similar triangles we find that d b = |AB| |AC| = √ 2 b b 2 + √ 2 b 2 ⇒ d = √ 2 b 2 √ 3b 2 = √ 6 3 b So h = b − d = b − √ 6 b = 3 − √ 6 3 3 b. So, using the formula V = πh 2 (r − h/3) from Exercise 6.2.51 with r = b,wefind that 3 − the volume of each of the caps is π √ 2 6 b b − 3 − √ 6 b = 15 − 6√ 6 · 6+√ 6 πb 3 = 2 − √ 7 3 3 · 3 9 9 3 27 6 πb 3 . So, using our first observation, the shared volume is V = 1 4 πb3 − 4 2 − √ 7 2 3 3 27 6 πb 3 = √ 28 27 6 − 2 πb 3 . 9. We can assume that the cut is made along a vertical line x = b>0, that the disk’s boundary is the circle x 2 + y 2 =1, and that the center of mass of the smaller piece (to the right of x = b)is 1 2 , 0 .Wewishtofind b to two decimal places. We have 1 1 2 = x = x · 2 √ 1 − x b 2 dx 1 2 √ . Evaluating the 1 − x b 2 dx numerator gives us − 1 1 (1 − b x2 ) 1/2 (−2x) dx = − 2 3 − x 2 3/2 1 = − 2 0 − 1 − b 2 3/2 = 2 (1 − b 2 3 3 ) 3/2 . b Using Formula 30 in the table of integrals, we find that the denominator is √ x 1 − x2 +sin −1 x 1 = √ 0+ π b 2 − b 1 − b 2 +sin −1 b . Thus, we have 1 2 2 = x = (1 − b 2 3 ) 3/2 π − b √ 1 − b 2 2 − sin −1 b ,or, equivalently, 2 (1 − b 2 ) 3/2 = π − 1 b √ 1 − b 3 4 2 2 − 1 2 sin−1 b. Solving this equation numerically with a calculator or CAS, we obtain b ≈ 0.138173,orb =0.14 m to two decimal places.
F. 380 ¤ CHAPTER 8 PROBLEMS PLUS TX.10 11. If h = L,thenP = area under y = L sin θ area of rectangle = π 0 L sin θdθ πL = [− cos θ]π 0 π = − (−1) + 1 π = 2 π . If h = L/2,thenP = area under y = 1 2 L sin θ area of rectangle = π 0 1 L sin θdθ 2 = [− cos θ]π 0 πL 2π = 2 2π = 1 π .
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