Chapter 5: Applications of Integration
Section 5.3: Volume by Slicing
Example 5.3.3
By the method of slicing, obtain the volume of the solid whose base in the xy-plane is the region bounded by the x-axis, and the curves y=sinx and x=π/2, and whose cross sections parallel to the yz-plane are equilateral triangles.
Solution
Mathematical Solution
Figure 5.3.3(a) contains an image of the solid. Figure 5.3.3(b) animates the slices.
use plots, plottools in
module()
local p1,p2,p3,p4,p5;
p1 := plot3d(2*y,x=0..Pi/2,y=0..sin(x)/2,color=red):
p2 := plot3d(2*(sin(x)-y),x=0..Pi/2,y=sin(x)/2..sin(x),color=blue):
p3 := plot3d(0,x=0..Pi/2,y=0..sin(x),color=green):
p4 := polygon([[Pi/2,0,0],[Pi/2,1,0],[Pi/2,1/2,1]],color=gray):
p5 := display([p1,p2,p3,p4],axes=frame, labels=[x,y,z],scaling=constrained,style=surface,tickmarks=[[0],0,0],orientation=[-50,75]);
print(p5);
end module:
end use:
Figure 5.3.3(a) The solid
local f,p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,T,k;
f := transform((x,y)->[x,y,0]):
T := x->display(polygon([[x,0,0],[x,sin(x),0],[x,sin(x)/2,sqrt(3)*sin(x)/2]],color=gray),transparency=.4):
p1 := plot(sin(x),x=0..Pi/2,y=0..1,color=black,thickness=2):
p2 := plot([[Pi/2,0],[Pi/2,1]],style=line, color=black,thickness=2):
p3 := plot([[0,0],[Pi/2,0]],style=line,color=black,thickness=2):
p4 := plot(sin(x),x=0..Pi/2,filled=true,color=green,transparency=.7):
p5 := display([p1,p2,p3,p4],scaling=constrained,tickmarks=[spacing(Pi/2,0),2]):
p6 := display([seq(T(k*Pi/2/30),k=0..30)],insequence=true):
p7:=spacecurve([x,sin(x)/2,sin(x)*sqrt(3)/2],x=0..Pi/2,color=black,thickness=2):
p8:=spacecurve([[Pi/2,0,0],[Pi/2,1/2,sqrt(3)/2]],numpoints=2,color=black,thickness=2):
p9:=spacecurve([[Pi/2,1/2,sqrt(3)/2],[Pi/2,1,0]],numpoints=2,color=black,thickness=2):
p10 := display([p6,f(p5),p7,p8,p9],axes=frame,labels=[x,y,z],orientation=[-50,70],tickmarks=[[0],0,0],view=0..1);
print(p10);
Figure 5.3.3(b) Animation of slices
The generic slice, an equilateral triangle, is sketched in Figure 5.3.3(c). The length of the base of this triangle is necessarily sinx, so the area of the triangle is Ax=12 sin2x3/2, obtained with the general formula for the area of triangle ABC, namely, 12 a⋅b⋅sinC, where a and b are the lengths of the sides forming the angle at vertex C.
Alternatively, the area is half the product of base times height h, where h is the length of the altitude (dotted red line in Figure 5.3.3(c)).
Either half is a 30-60-90 right triangle from which h is easily obtained by similarity of triangles.
hsinx2=31⇒h=32sinx
In either event, the volume of the solid is then
V=∫0π/2Ax ⅆx = 34∫0π/2sin2x ⅆx=π 316
local m1,m2,m3,m4;
m1 := polygon([[0,0],[1,0],[1/2,sqrt(3)/2]],style=line,color=black):
m2 := plot([[1/2,0],[1/2,sqrt(3)/2]],style=line, linestyle=2,color=red):
m3 := textplot({[.65,.2,typeset(h=sqrt(3)*sin(x)/2)], [.3,.07,typeset(sin(x)/2)], [.75,.07,typeset(sin(x)/2)]}):
m4 := display([m1,m2,m3],scaling=constrained,tickmarks=[[0],0],labels=[y,z]);
print(m4);
Figure 5.3.3(c) Generic slice - equilateral triangle
Maple Solution
Expression palette: Definite-integral template
Context Panel: Evaluate and Display Inline
34∫0π/2sin2x ⅆx = 1163π
Stepwise evaluation of the integral
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Control-drag just the integral.
Context Panel: Student Calculus1≻All Solution Steps
∫0π/2sin2x ⅆx→show solution stepsIntegration Steps∫0π2sinx2ⅆx▫1. Rewrite◦Equivalent expressionsinx2=12−cos2x2This gives:∫0π212−cos2x2ⅆx▫2. Apply the sum rule◦Recall the definition of the sum rule∫fx+gxⅆx=∫fxⅆx+∫gxⅆxfx=12gx=−cos2x2This gives:∫0π212ⅆx+∫0π2−cos2x2ⅆx▫3. Apply the constant rule to the term ∫12ⅆx◦Recall the definition of the constant rule∫Cⅆx=Cx◦This means∫12ⅆx=x2We can now rewrite the integral as:π4+∫0π2−cos2x2ⅆx▫4. Apply the constant multiple rule to the term ∫−cos2x2ⅆx◦Recall the definition of the constant multiple rule∫Cfxⅆx=C∫fxⅆx◦This means:∫−cos2x2ⅆx=−∫cos2xⅆx2We can rewrite the integral as:π4−∫0π2cos2xⅆx2▫5. Apply a change of variables to rewrite the integral in terms of u◦Let u beu=2x◦Isolate equation for xx=u2◦Differentiate both sidesdx=du2◦Substitute the values for x and dx back into the original∫0π2cos2xⅆx=∫0πcosu2ⅆuThis gives:π4−∫0πcosu2ⅆu2▫6. Apply the constant multiple rule to the term ∫cosu2ⅆu◦Recall the definition of the constant multiple rule∫Cfuⅆu=C∫fuⅆu◦This means:∫cosu2ⅆu=∫cosuⅆu2We can rewrite the integral as:π4−∫0πcosuⅆu4▫7. Evaluate the integral of cos(u)◦Recall the definition of the cos rule∫cosuⅆu=sinu◦Apply limits of definite integralsinuu=π|sinuu=π−sinuu=0|sinuu=0This gives:π4
This value of the integral, namely π/4, must be multiplied by 3/4 to obtain 3π/16.
Of course, the stepwise evaluation of the integral can be implemented interactively by means of the tutor. Additionally, the set of Integration Rules in the Integration Methods tutor can be accessed through the Context Panel, as per Figure 5.3.3(d).
Figure 5.3.3(d) Context Panel access to Integration Rules
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