Chapter 5: Applications of Integration
Section 5.2: Volume of a Solid of Revolution
Example 5.2.2
If A is the plane region bounded by the x-axis and the graphs of y=x2 and x=1, use the method of disks to calculate the volume of the solid of revolution formed when A is rotated about the line y=−1.
Solution
Mathematical Solution
Figures 5.2.2(a-c) illustrate the essential steps in the method of disks as applied to this example. In Figure 5.2.2(a) the region A is shaded, with the arrows representing the radii of rotation. The black arrow corresponds to the outer radius R=1+x2; the green arrow, to the inner radius r=1.
module() local p1,p2,VR,Vr,p3,p4,p5; p1:=plot(x^2,x=0..1,filled=[color=brown,transparency=.4],color=black,labels=[x,y],tickmarks=[2,3],thickness=3): p2:=plot(-1,x=0..1,color=black,thickness=2): VR:=VectorCalculus:-RootedVector(root=[3/4,-1],<0,25/16>): Vr:=VectorCalculus:-RootedVector(root=[1/4,-1],<0,1>): p3:=VectorCalculus:-PlotVector([VR,Vr],color=[black,green],width=.03): p4:=plots:-textplot({[.95,-.3,typeset(R=1+x^2)],[.36,-.3,typeset(r=1)]},font=[default,12]): p5:=plots:-display(p3,p1,p2,p4); print(p5); end module:
Figure 5.2.2(a) Region A
Student:-Calculus1:-VolumeOfRevolution(x^2,0,0..1,axis=horizontal, distancefromaxis=-1,showvolume= true,showregion=true,output=plot,axes=frame,caption= "",volumeoptions=[color=red,transparency=0],scaling=constrained,tickmarks=[2,[-3,0,3],[-3,-2,-1,0,1]],labels=[x,z,y],orientation=[-150,85,-10]);
Figure 5.2.2(b) The solid
Student:-Calculus1:-VolumeOfRevolution(x^2,0,0..1,axis=horizontal, distancefromaxis=-1,showvolume=false,showsum=true,showregion=false, method =midpoint,partition=6,output=plot,axes=frame,sumvolumeoptions=[color= brown,transparency=0,lightmodel=light3],caption="",tickmarks=[2,[0],5],labels=[x,z,y],scaling=constrained,orientation=[-150,85,-10]);
Figure 5.2.2(c) Disks
The solid of rotation itself is shown in Figure 5.2.2(b). The bounding curve y=x2 is drawn on the surface of the solid. Note how the z-axis is out of the xy-plane, which is the plane of the viewing screen. Figure 5.2.2(c) shows the solid sliced into a stack of disks. Each such disk has a hole, so the punctured disk resembles a washer. The inner radius of the washer is r=1; the outer, R=1+x2.
One washer has volume π R2−r2 dx, leading to the definite integral listed in Table 5.2.1.
The actual volume, computed as per Table 5.2.1, is π ∫01x4+2 x2 ⅆx = 13π15
Maple Solution
Figure 5.2.2(d) shows the Volume of Revolution tutor applied to the given solid.
Note the inclusion of the bounding function gx=0, without which the volume would be incorrectly computed as 28 π/15.
The Plot Options button has been used to change the axes style (frame) and to set Constrained Scaling.
Because Maple can't determine which of R or r is greater, the absolute value of the difference R2−r2 is integrated.
Using the Calculus palette's definite-integral template, the volume of the solid of revolution (computed by the methods of disks) is
π ∫01x4+2 x2 ⅆx = 13π15
Figure 5.2.2(d) tutor
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