Transcript Chapter 6
Section 6.1 Volumes By Slicing and Rotation About an Axis Generalized Cylinder A cylinder is a solid that is generated when a plane region is translated along a line or axis that is perpendicular to the region. If a cylindrical solid is generated y translating a region of area A through a distant h, then h is called the height of the cylinder, and the volume V of the cylinder is defined to be V=Ah =[area of a cross section] X [height] Volumes By Slicing • To solve this problem, we begin by dividing the interval [a, b] into n subintervals, thereby dividing the solid into n slabs. • If we assume that the width of the kth subinterval is xk, then the volume of the kth slab can be approximated by the volume A(xk)xk of a right cylinder of width (height) xk and cross-sectional area A(xk), where xk is a point in the kth subinterval. • Adding these approximations yields the following Riemann sum that approximates the volume V: n V A( xk )xk k 1 • Taking the limit as n increases and the width of the subintervals approach zero yields the definite integral V lim max xk 0 n A( x )x k 1 k k b a A( x)dx Volume Formula The Volume of a solid can be obtained by integrating the cross-sectional area from one end of the solid to the other. How to Calculate the Volume To apply the formula in the definition to calculate the volume of a solid, take the following steps: Example A pyramid 3m high has a square base that is 3 m on a side. The cross section of the pyramid perpendicular to the altitude x m down from the vertex is a square x m on a side. Find the volume of the pyramid. Solids of Revolution: The Disk Method Examples Example: The region between the curve y x , 0 x 4, and the x-axis is revolved about the x-axis to generate a solid. Find its volume. Example Example: Find the volume of the solid generated by revolving the region bounded by y x and the lines y=1, x=4 about the line y=1. To find the volume of a solid generated by revolving a region between the y-axis and a curve x=R(y), c y d, about the y-axis, we use the same method with x replaced by y. In this case, the circular crosssection is A(y)= [radius]2 = [R(y)]2 Example Example: Find the volume of the solid generated by revolving the region between the y-axis and the curve x=2/y, 1 y 4, about the y-axis. Example Example: Find the volume of the solid generated by revolving the region between the parabola x=y2+1 and the line x=3 about the line x=3. Solids of Revolution: The Washer Method Example: The region bounded by the curve y=x2+1 and the line y=-x+3 is revolved about the x-axis to generate a solid. Find the volume of the solid. Example Example: The region bounded by the parabola y=x2 and the line y=2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid. Section 6.2 Volumes by Cylindrical Shells The method of slicing in section 6.1 is sometimes awkward to apply. To overcome This difficulty, we use the same integral definition for volume, but obtain the area by slicing through the solid in a different way. Volume of Cylindrical Shells A cylindrical shell is a solid enclosed by two concentric right circular cylinders. The volume V of a cylindrical shell with inner radius r1, outer radius r2, and height h can be written as V=2 [1/2(r1+r2) ] h (r2-r1) So V=2 [ average radius ] [height] [thickness] Method of Cylindrical Shells • The idea is to divide the interval [a, b] into n subintervals, thereby subdividing the region R into n strips, R1, R2,, …, Rn. • When the region R is revolved about the y-axis, these strips generate “tube-like” solids S1, S2, …, Sn that are nested one inside the other and together comprise the entire solid S. • Thus the volume V of the solid can be obtained by adding together the volumes of the tubes; that is V=V(S1)+V(S2)+…+V(Sn). Method of Cylindrical Shells Method of Cylindrical Shells Suppose that the kth strip extends from xk-1 to xk and that the width of the strip is xk. If we let xk* be the midpoint of the interval [xk-1, xk], and if we construct a rectangle of height f(xk*) over the interval, then revolving this rectangle about the y-axis produces a cylindrical shell of average radius xk*, height f(xk*), and thickness xk. Then the volume Vk of this cylindrical shell is Vk=2xk*f(xk*) xk n V 2 xk* f ( xk* )xk k 1 Hence, we have V lim max xk 0 n 2 x k 1 * k b f ( x )xk 2 xf ( x)dx * k a Volume by Cylindrical Shells about the y-axis Let f be continuous and nonnegative on [a, b] (0a<b), and let R be the region that is bounded above by y=f(x), below by the x-axis, and on the sides by the lines x=a and x=b. Then the volume V of the solid of revolution that is generated by revolving the region R about the y-axis is given by b V 2 xf ( x)dx a Generally Summary Example Example: The region bounded by the curve y x , the x-axis, and the line x=4 is revolved about the y-axis to generate a solid. Find the volume of the solid. Example So far, we have used vertical axes of revolution. For horizontal axes, we replace the x’s with y’s. Example: The region bounded by the curve y x , the x-axis, and the line x=4 is revolved about the x-axis to generate a solid. Find the volume of the solid by the shell method. Examples Example. Use cylindrical shells to find the volume of the solid generated when the region enclosed between y x , x 1, x 4, and the x-axis is revolved about the y-axis. Solution: 6.3 Length of a Plane Curve Arc Length Formula for Parametric Curves Example: Find the circumference of a circle of radius 2 from the parametric equations x 2cos t , y 2sin t Solution: (0 t 2 ) Example Example: Find the length of the astroid. x cos3 t, y sin3 t, (0 t 2 ) y=f(x) is a smooth curve on [a, b] if f ’ is continuous on [a, b]. Where convenient, (3) can also be expressed as Example Find the arc length of the curve y x 3/2 Solution: from (1, 1) to (2,2 2) Dealing with Discontinuities in dy/dx At a point on a curve where dy/dx fails to exist, dx/dy may exist and we may be able to Find the curve’s length by expressing x as a function of y and applying the following: Example Example: Find the length of the curve y=(x/2)2/3 from x=0 to x=2.