Chapter 5: Applications of Integration
Section 5.5: Surface Area of a Surface of Revolution
Example 5.5.5
Calculate the surface area of the surface of revolution formed when the graph of y=e−x,x∈0,2, is rotated about the y-axis.
Solution
Mathematical Solution
According to Table 5.5.1, the "formula" 2 π ∫abρ ⅆs becomes
2 π ∫02x 1+ⅆⅆ x e−x2 ⅆx = 2 π ∫02x 1+e−2 x ⅆx ≐ 13.23912130
Maple cannot provide an antiderivative for the integrand x 1+e−2 x, so the definite integral is evaluated numerically. While a Riemann sum with approximating rectangles could be used, more efficient methods will be seen in Chapter 6.
Maple Solution
In Figure 5.5.5(a) the tutor has been applied to the graph of y=e−x rotated about the y-axis. Note how e−x is entered into the tutor as exp−x. Moreover, to calculate with the exponential function, the exponential e must be entered from a palette (e.g., Common Symbols) or via command completion (Tools menu, or the Escape key).
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 not available, so the tutor provided just a floating-point approximation.
Figure 5.5.5(a) Surface of Revolution tutor
Interactive evaluation of the surface-area integral
Expression palette: Definite-integral and derivative templates Press the Enter key.
Context Panel: Approximate≻10 (digits)
2 π∫02x 1+ⅆⅆ x ⅇ−x2 ⅆx
2π∫02x1+ⅇ−x2ⅆx
→at 10 digits
13.23912130
Application of the SurfaceOfRevolution command
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
SurfaceOfRevolutionⅇ−x,x=0..2,axis=vertical, output=integral
∫022πx1+ⅇ−x2ⅆx
SurfaceOfRevolutionⅇ−x,x=0..2,axis=vertical = ∫022πx1+ⅇ−x2ⅆx→at 10 digits13.23912130
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