Chapter 5: Applications of Integration
Section 5.3: Volume by Slicing
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Example 5.3.2
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By the method of slicing, obtain the volume of the solid whose base is an equilateral triangle of side , and whose cross sections are squares.
In particular, the equilateral triangle lies in the plane , has a vertex at the origin, and an altitude along the -axis. The square cross sections are parallel to the -plane.
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Solution
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Mathematical Solution
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Figure 5.3.2(a) contains an image of the solid. Figure 5.3.2(b) animates the slices.
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use plots, plottools in
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local a,q1,q2,q3,q4,q5,q6;
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q1 := polygon([[0,0,0],[a,-1,0],[a,1,0]]):
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q2 := polygon([[0,0,0],[a,1,0],[a,1,2]]):
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q3 := polygon([[0,0,0],[a,-1,0],[a,-1,2]]):
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q4 := polygon([[0,0,0],[a,-1,2],[a,1,2]]):
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q5 := polygon([[a,-1,0],[a,1,0],[a,1,2],[a,-1,2]]):
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q6 := display([q1,q2,q3,q4,q5],scaling=constrained,axes=frame, labels=[x,y,z],tickmarks=[[0],[0],0],orientation=[-120,75]);
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Figure 5.3.2(a) The solid
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use plots, plottools in
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local a,p1,p2,p3,p4,p5,p6,p7,p8,pp;
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p1 := proc(u)
display(polygon([[u,u/a,0],[u,-u/a,0],[u,-u/a,2*u/a],[u,u/a,2*u/a]], color=red),transparency=.7):
end proc:
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p2 := spacecurve([[0,0,0],[1,1/a,2/a]],color=black):
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p3 := spacecurve([[0,0,0],[1,-1/a,2/a]],color=black):
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p4 := spacecurve([[0,0,0],[1,1/a,0]],color=black):
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p5 := spacecurve([[0,0,0],[1,-1/a,0]],color=black):
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p6 := polygon([[1,-1/a,0],[1,1/a,0],[1,1/a,2/a],[1,-1/a,2/a]],style=line,color=black):
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p7 := polygon([[0,0,0],[1,1/a,0],[1,-1/a,0]],color=yellow):
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pp:=display([p2,p3,p4,p5,p6,p7]):
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p8 := animate(p1,[u],u=0..1,frames=41,axes=none,orientation=[-135,80],labels=[x,y,z],tickmarks=[1,1,1],scaling=constrained,background=pp,paraminfo=false);
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Figure 5.3.2(b) Animation of slices
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The large triangle in Figure 5.3.2(c) is a schematic of the base equilateral triangle of side . The figure is used to relate the height of a square slice (see Figure 5.3.2(b)) to the length of a side of this slice. When the slice is located at , the base of the square slice is , so the area of that square is .
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The slices start at and end at , where is the altitude of the equilateral triangle of side .
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The yellow triangle in Figure 5.3.2(c) is a 30-60-90 right triangle, which is therefore similar to a right triangle with sides . (See the red numbers in Figure 5.3.2(c).)
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Similar triangles give the following proportions:
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and
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Consequently, , and
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use plots, plottools in
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local a,p1,p2,p3,p4,p5;
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p1 := polygon([[0,0],[1,-1/a],[1,1/a]],style=line):
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p2 := polygon([[0,0],[1/2,0],[1/2,1/2/a]],color=yellow):
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p3 := textplot({[.3,.03,typeset(x)],[.47,.13,typeset(y)], [.6,.4,typeset(s)], [.098,.027,typeset(alpha)], [.95,.25,typeset(s/2)], [1.05,.034,typeset(x)], [.56,-.05,typeset(h)]}):
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p4 := textplot({[.53,.13,typeset(1)],[.25,.19,typeset(2)],[.27,-.043,typeset(sqrt(3))]},color=red):
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p5 := display([p1,p2,p3,p4],scaling=constrained,tickmarks=[[0],0],labels=[" ",y],view=[0..1.1,-.6..0.6]);
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Figure 5.3.2(c) Schematic of base triangle
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Maple Solution
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Expression palette: Definite-integral template
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Context Panel: Evaluate and Display Inline
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