a. Explanation Given information: A solid is bounded by two bases in the horizontal plates z = h 2 and z = − h 2 and by a surface with the property that every horizontal cross section has area given by an expression of the form A = a 0 z 3 + a 1 z 2 + a 2 z + a 3 Formula used: A solid S with cross-sectional area A ( x ) perpendicular to the x-axis at each point on the interval [ a , b ] has volume V = ∫ a b A ( x ) d x . Proof: The cross-sectional area is A = a 0 z 3 + a 1 z 2 + a 2 z + a 3 . Then the volume is V = ∫ h 2 − h 2 A ( z ) d z . V = ∫ h 2 − h 2 ( a 0 z 3 + a 1 z 2 + a 2 z + a 3 ) d z = a 1 h 3 12 + a 3 h The base areas and M are given by B 1 = a 0 h 3 8 + a b. To determine To prove: The formulas for the volume of the cone and a sphere can be obtained from the formula V = 1 6 h ( B 1 + B 2 + 4 M ) for certain special choices a 0 , a 1 , a 2 a n d a 3 .

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ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 6, Problem 96SP

To determine

To prove:The solid has volume V=16h(B1+B2+4M) where B1 and B2 are the area of the bases and M is the area of the middle horizontal cross section at z=0.

To determine

To prove:The formulas for the volume of the cone and a sphere can be obtained from the formula V=16h(B1+B2+4M) for certain special choices a0,a1,a2 and a3 .

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Chapter 6, Problem 95SP

Chapter 6, Problem 97SP

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To prove: The solid has volume V = 1 6 h ( B 1 + B 2 + 4 M ) where B 1 and B 2 are the area of the bases and M is the area of the middle horizontal cross section at z = 0.

Solution Summary: The author explains that a solid with cross-sectional area is perpendicular to the x-axis at each point on the interval left[a,bright].

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To prove:The solid has volume V=16h(B1+B2+4M) where B1 and B2 are the area of the bases and M is the area of the middle horizontal cross section at z=0.

To prove:The formulas for the volume of the cone and a sphere can be obtained from the formula V=16h(B1+B2+4M) for certain special choices a0,a1,a2 and a3 .

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Chapter 6 Solutions