Chapter 5: Double Integration
Section 5.7: Double Integration in Polar Coordinates
Example 5.7.7
Calculate the area that is inside the circle r=4 sinθ but outside the circle r=2.
Solution
Mathematical Solution
Figure 5.7.7(a) shows the circle r=4 sinθ (in black), the limaçon r=2 (in red), and the region whose area is to be calculated, (in green).
Figure 5.7.7(b) is an animation in which the two curves are drawn under the action of the slider in the animation toolbar. The polar angle appears above the vertical axis. Use this animation to infer the appropriate angles for the iterated double integral that follows.
Inspired by the animation in Figure 5.7.7(b), Figure 5.7.7(c) is a graph of r=4 sinθ and r=2. It suggests that the two curves intersect at θ=π/6 and 5 π/6. The shaded region in Figure 5.7.7(a) is traced for θ∈π/6,5 π/6.
use plots in module() local R,R1,R2,p1,p2,p3; R:=sqrt(x^2+y^2); R1:=4*sin(t); R2:=2; p1:=plot([R1,R2],t=0..2*Pi,coords=polar,color=[black,red],thickness=[1,3]); p2:=inequal([R>=2,R<=4*y/R],x=-2..2,y=1..4,color=green); p3:=display(p1,p2,scaling=constrained,labels=[x,y]); print(p3); end module: end use:
Figure 5.7.7(a) Region
Figure 5.7.7(b) Animation
Figure 5.7.7(c) Intersections
The area of the region shaded in green in Figure 5.7.7(a) is
∫π/65 π/6∫24 sinθr ⅆr ⅆθ = 23+4 π/3
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Access the MultiInt command via the Context Panel
Type the integrand, 1.
Context Panel: Student Multivariate Calculus≻Integrate≻Iterated Fill in the fields of the two dialogs shown below. Note that the range for r is not completely visible.
Context Panel: Evaluate Integral
1→MultiInt∫16π56π∫24sinθrⅆrⅆθ=23+43π
The task template in Table 5.7.7(a) can be used to visualize the region of integration over which a given iterated integral acts. Select an order of integration, and provide an integrand of 1 to compute area. Supply the limits of integration, and use the Exact button to obtain the value of the iterated integral, and the Draw Graphs button to obtain the two figures provided by the task template.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Polar
Evaluate ∬RΨr,θ dA and Graph R
Area Element dA
r dr dθ
r dθ dr
, Ψ=
Value of Integral
G=
b=
g=
a=
Bounding Curves
"Volume"
Table 5.7.7(a) Task template for visualizing a polar region of integration
The figure on the left is an animation that shows how the radial cone representing dθ traverses the region of integration. The figure on the right is a representation of the volume of a solid of height 1, with base the region of integration. Since the height is 1, the number computed for the volume is the same number as the area. If this figure is rotated and viewed from above, it appears to be a shaded version of the region of integration. These visual clues help to decide of the polar area has been properly identified and calculated.
The details of an interactive calculation of the required area appear in Table 5.7.7(b).
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Find the zeros of 4 sinθ=2
Write the equation.
Context Panel: Student Calculus1≻Solve≻Find Roots Complete the dialog as per Figure 5.7.7(d)
Figure 5.7.7(d) Roots dialog
4 sinθ=2→roots16π,56π
Implement and evaluate the iterated integral
Calculus palette: Iterated double integral template
Context Panel: Evaluate and Display Inline
∫π/65 π/6∫24 sinθr ⅆr ⅆθ = 23+43π
Table 5.7.7(b) Details of the interactive calculation of the required area.
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