# X 1if we we take a cross section dark purple line our

• Peterliao
• 244

This preview shows page 57 - 62 out of 244 pages.

##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
The document you are viewing contains questions related to this textbook.
Chapter 7 / Exercise 1
Calculus
Stewart
Expert Verified
=0x1.If we we take a cross section (dark purple line) our our plot androtate around the x-axis, we obtain the shape shown above(a hollow cone).If we look down the cone, we will see the cross sections which arecalledwashers.For each cross section, we haveRadiusinner=x2Radiusouter=x57
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
The document you are viewing contains questions related to this textbook.
Chapter 7 / Exercise 1
Calculus
Stewart
Expert Verified
Janelle Resch’s MATH 138 Notes Spring 2014So to find the cross-sectional area we haveA(x) =A(x)outer-A(x)inner=π(radiusouter)2-π(radiusinner)2=π(x)2-π(x2)2=πx2-πx4Thus, by definition, the volume of the solid between 0x1 isV=Z10A(x) dx=πZ10x2-x4dx=π13x3-15x510=π13-15=2π15.Remark:So far, all the examples we have looked at are calledsolids ofrevolution. These solids are given this name because they allrevolve around anaxis of rotation(i.e., a line).Volumes by the Method of Cylindrical ShellsWhen we try to determine the volume for more complicated curves andhence solids, the method of slicing becomes more difficult. In such cases, wecan use themethod of cylindrical shells.Let’s say we have a cylindrical shellwith in inner radius,denoted byr1, and an outer radius, denotedbyr2.Let r be the average ra-dius.The height of the cylinders ishandlet Δrbe the thickness of the shell.58
Janelle Resch’s MATH 138 Notes Spring 2014The volume of the cylindrical shell is determined byV=Vouter-Vinnter=π(r2)2h-π(r1)2h=πh(r22-r12)=πh(r2-r1|{z}Δr)(r2+r1)22= 2πhΔrr2+r12|{z}average radius of shell= 2πhrΔrOr in other words, the volume V, of a cylindrical shell isV= (circumference)(height)(thickness).Now, let’s consider a functiony=f(x), wheref(x)0 andaxb.Let S be the solid obtained by rotatingf(x) about the y-axis.Next, divide S into n subintervals, denoted bydV. The width of each subin-terval is Δx, where we will denote the midpointx. We know that the volumeof each subinterval isdV= (circumference)(height)(thickness)dV= 2πx f(x) Δx59
Janelle Resch’s MATH 138 Notes Spring 2014To approximate the volume V of S, we want to sum the volumes of all theshellsVnXi=12πxif(xi) Δxi.To obtain the best possible approximation, we taken→ ∞, we have theRiemann sumV= limn→∞nXi=12πxif(xi) Δxi.And the limit of the Riemann sums is defined as the definite integral, henceV=Zba2πx f(x) dx.Example:The region R enclosed by the curvesy=xandy=x2is rotatedabout the y-axis. Find the volume of the resulting solid.Notice the curvesy=xandy=x2intersect at (0,0) and (1,1).=0x1.If we we take a cross section (dark purple line) our our plot androtate around the y-axis, we obtain the shape shown above60
Janelle Resch’s MATH 138 Notes Spring 2014For each cross section, we haveheight =x-x2Radius =xThus, the volume between 0x1 is defined asV=Z10(circumference)(height) dx=Z10(2πx)(x-x2) dx= 2πZ10x2-x3dx= 2π13x3-14x410= 2π13-14=π6.