(a) Show that the volume of the solid in R that lies below the surface z = f(x, y) and above the region D is equal to the volume of a cylinder of radius R and height z. (Stop and think: Does this assertion make intuitive sense? (You do not have to submit your answer to this question.)]

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
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The average value z of z = f(x, y) on a subset D C R? is defined to be
1
/, f(x, y) dA
area(D)
(provided the quantities on the RHS are well-defined and finite).
In what follows let R>0 be a fixed constant, let D = {(x, y): x2 + y? < R²}, and let f be a
continuous function that is defined on D and satisfies f(x, y) 2 0 for all (x, y) E D.
(a) Show that the volume of the solid in R³ that lies below the surface z = f(x, y) and
above the region D is equal to the volume of a cylinder of radius R and height z.
(Stop and think: Does this assertion make intuitive sense? (You do not have to submit
your answer to this question.)]
(b) Consider now the function f(x, y) =
V
R² – x² – y².
(i) Show that the average value of z = f(x, y) on D is z =R.
(ii) Describe in words the surface z = f(x, y). Be sure to give some justification for
your description.
(iii) Using your work above together with part (a), show that the volume of a ball of
radius R is TR.
(Note: In c. 225 BCE, Archimedes showed that the volume of a ball of radius R
is equal to nR³ by showing that it is equal to two-thirds the volume of the smallest
cylinder that surrounds it. He did so using an ingenious argument that involved an early
form of integral calculus, almost 2000 years before the "official" discovery of calculus by
Newton and Leibniz. Our approach above is very close in spirit (though not identical
to) Archimedes's method.]
Transcribed Image Text:The average value z of z = f(x, y) on a subset D C R? is defined to be 1 /, f(x, y) dA area(D) (provided the quantities on the RHS are well-defined and finite). In what follows let R>0 be a fixed constant, let D = {(x, y): x2 + y? < R²}, and let f be a continuous function that is defined on D and satisfies f(x, y) 2 0 for all (x, y) E D. (a) Show that the volume of the solid in R³ that lies below the surface z = f(x, y) and above the region D is equal to the volume of a cylinder of radius R and height z. (Stop and think: Does this assertion make intuitive sense? (You do not have to submit your answer to this question.)] (b) Consider now the function f(x, y) = V R² – x² – y². (i) Show that the average value of z = f(x, y) on D is z =R. (ii) Describe in words the surface z = f(x, y). Be sure to give some justification for your description. (iii) Using your work above together with part (a), show that the volume of a ball of radius R is TR. (Note: In c. 225 BCE, Archimedes showed that the volume of a ball of radius R is equal to nR³ by showing that it is equal to two-thirds the volume of the smallest cylinder that surrounds it. He did so using an ingenious argument that involved an early form of integral calculus, almost 2000 years before the "official" discovery of calculus by Newton and Leibniz. Our approach above is very close in spirit (though not identical to) Archimedes's method.]
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