Chapter 5: Applications of Integration
Section 5.3: Volume by Slicing
Example 5.3.2
By the method of slicing, obtain the volume of the solid whose base is an equilateral triangle of side s, and whose cross sections are squares.
In particular, the equilateral triangle lies in the plane z=0, has a vertex at the origin, and an altitude along the x-axis. The square cross sections are parallel to the yz-plane.
Solution
Mathematical Solution
Figure 5.3.2(a) contains an image of the solid. Figure 5.3.2(b) animates the slices.
use plots, plottools in
module()
local a,q1,q2,q3,q4,q5,q6;
a := sqrt(3):
q1 := polygon([[0,0,0],[a,-1,0],[a,1,0]]):
q2 := polygon([[0,0,0],[a,1,0],[a,1,2]]):
q3 := polygon([[0,0,0],[a,-1,0],[a,-1,2]]):
q4 := polygon([[0,0,0],[a,-1,2],[a,1,2]]):
q5 := polygon([[a,-1,0],[a,1,0],[a,1,2],[a,-1,2]]):
q6 := display([q1,q2,q3,q4,q5],scaling=constrained,axes=frame, labels=[x,y,z],tickmarks=[[0],[0],0],orientation=[-120,75]);
print(q6);
end module:
end use:
Figure 5.3.2(a) The solid
local a,p1,p2,p3,p4,p5,p6,p7,p8,pp;
a := sqrt(3);
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:
p2 := spacecurve([[0,0,0],[1,1/a,2/a]],color=black):
p3 := spacecurve([[0,0,0],[1,-1/a,2/a]],color=black):
p4 := spacecurve([[0,0,0],[1,1/a,0]],color=black):
p5 := spacecurve([[0,0,0],[1,-1/a,0]],color=black):
p6 := polygon([[1,-1/a,0],[1,1/a,0],[1,1/a,2/a],[1,-1/a,2/a]],style=line,color=black):
p7 := polygon([[0,0,0],[1,1/a,0],[1,-1/a,0]],color=yellow):
pp:=display([p2,p3,p4,p5,p6,p7]):
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);
print(p8);
Figure 5.3.2(b) Animation of slices
The large triangle in Figure 5.3.2(c) is a schematic of the base equilateral triangle of side s. 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 x, the base of the square slice is 2 y, so the area of that square is Ax=2 y2.
The slices start at x=0 and end at x=h, where h is the altitude of the equilateral triangle of side s.
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 1,2,3. (See the red numbers in Figure 5.3.2(c).)
Similar triangles give the following proportions:
yx=13⇒y=x3
and
13=s/2h⇒h=s 32
Consequently, Ax=2 x/32, and
local a,p1,p2,p3,p4,p5;
p1 := polygon([[0,0],[1,-1/a],[1,1/a]],style=line):
p2 := polygon([[0,0],[1/2,0],[1/2,1/2/a]],color=yellow):
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)]}):
p4 := textplot({[.53,.13,typeset(1)],[.25,.19,typeset(2)],[.27,-.043,typeset(sqrt(3))]},color=red):
p5 := display([p1,p2,p3,p4],scaling=constrained,tickmarks=[[0],0],labels=[" ",y],view=[0..1.1,-.6..0.6]);
print(p5);
Figure 5.3.2(c) Schematic of base triangle
V=∫0hAx ⅆx = ∫0s3/22 x/32 ⅆx=36s3
Maple Solution
Expression palette: Definite-integral template
Context Panel: Evaluate and Display Inline
∫0s3/22 x/32 ⅆx = 16s33
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