Chapter 5: Applications of Integration
Section 5.5: Surface Area of a Surface of Revolution
Example 5.5.3
Calculate the surface area of the surface of revolution formed when the graph of y=x2,x∈0,1, is rotated about the y-axis.
Solution
Mathematical Solution
According to Table 5.5.1, the "formula" 2 π ∫abρ ⅆs becomes
2 π ∫01x1+ⅆⅆ x x22 ⅆx
=2 π∫01x 1+4 x2 ⅆx
=2 π112⁢4⁢x2+13/201
=2⁢π⁢512⁢5−112
=π 55−1/6
≐5.330413500
An antiderivative for x 1+4 x2 can be found, for example, by making the elementary substitution u=1+4 x2 so that du=8 x dx. Then, x dx=du/8 and the indefinite integral becomes
∫u1/28 ⅆu=18u3/23/2=1+4 x23/212
Maple Solution
In Figure 5.5.3(a) the tutor has been applied to the graph of y=x2 rotated about the y-axis. The Plot Options section has been used to impose one-to-one scaling, and the frame axis style. An exact representation of the value of the surface-area integral is returned, along with a floating-point equivalent. Note the SurfaceOfRevolution command at the bottom of the tutor. This command can return the graphs shown in Figures 5.5.3(a-b), the unevaluated integral, or the value of the surface area.
Figure 5.5.3(a) Surface of Revolution tutor
Student:-Calculus1:-SurfaceOfRevolution(x^2,x=0..1,axis=vertical,partition=3,showsum=true, showsurface=false,output= plot,caption="",axes=frame,orientation=[-105,80,0],labels=[x,z,y],tickmarks=[[-1,0,1],[0],2],sumsurfaceoptions=[color=red],showfunction=false,scaling=constrained);
Figure 5.5.3(b) Segmentation into frustums
Figure 5.5.3(b) shows the surface of revolution segmented into three frustums, an image that can be generated in the tutor by selecting "Frustums" in the Display section, and then pressing the Display button. The default number of subintervals is 6, but this has been changed to 3 in the figure. In addition to the graph in Figure 5.5.3(b), the tutor also provides the following midpoint Riemann approximating sum.
π⁢∑i=1319⁢13⁢i+13⁢i+13⁢4⁢i2-4⁢i+10≐8.321345866
It takes a considerably greater number of segments for the midpoint Riemann sum to give a better approximation to the actual value of the surface area.
Interactive evaluation of the surface-area integral
Expression palette: Definite-integral and derivative templates
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻10 (digits)
2 π∫01x1+ⅆⅆ x x22 ⅆx = 2⁢π⁢512⁢5−112→at 10 digits5.330413500
Application of the SurfaceOfRevolution command
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
SurfaceOfRevolutionx2,x=0..1,axis=vertical, output=integral = ∫012⁢π⁢x⁢4⁢x2+1ⅆx
SurfaceOfRevolutionx2,x=0..1,axis=vertical = 16⁢π⁢5⁢5−1
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