Chapter 7: Additional Applications of Integration
Section 7.2: Integration in Polar Coordinates
Example 7.2.8
Working in polar coordinates, calculate the area common to the circle r1=sinθ and the inner loop of the limaçon r2=1/5+cosθ.
Solution
Mathematical Solution
Figure 7.2.8(a) shows the intersection of circle and the limaçon. The region whose area is to be calculated is divided into two subregions designated R1 and R2.
The (blue) dividing line connects the origin O with the intersection point P, whose Cartesian coordinates are 12/25,9/25=0.48,0.36.
As per Figure 7.1.11(a), the limaçon is drawn with θ∈−π,π; the circle, with θ∈0,π.
At θ=−π, the limaçon starts at 4/5,0 on the x-axis, arrives at P when θ=arctan3/4−π, and at the origin when θ=arccos1/5−π.
use plots in module() local p1,p2,p3,p4,r1,r2; r1:=sin(t); r2:=1/5+cos(t); p1:=implicitplot([r=r1,r=r2],r=-2..2,t=-Pi..Pi,coords=polar,gridrefine=3,scaling=constrained,color=[black,red]): p2:=plot([[0,0],[12/25,9/25]],style=line, color=blue): p3:=textplot({[.55,.4,typeset(P)],[.21,.24,typeset(R[1])],[.27,.14,typeset(R[2])]}): p4:=plot([[12/25,9/25]],style=point,symbol=solidcircle,symbolsize=15,color=green): print(display(p1,p2,p3,p4,scaling=constrained,labels=[x,y],tickmarks=[[0,.8,1.2],[-.6,0,1]])); end module: end use:
Figure 7.2.8(a) Intersection of circle and limaçon
The area of region R1 is given by the integral
A1
=12∫arctan3/4−πarccos1/5−π1/5+cosθ2 ⅆθ
=27100arccos15−arctan34−3650
≐0.049
On the circle, the arc bounding region R2 is drawn with θ∈0,arcsin3/5. Hence, the area of region R2 is given by the integral
A2
=12∫0arcsin3/5sin2θ ⅆθ
=arcsin3/54−325
≐0.041
The total area of the region common to the circle and the inner loop of the limaçon is then the sum A1+A2≐0.090.
The astute reader will realize that the essence of this calculation is the determination of the bounding angles defining regions R1 and R2, wherein the difficulty is the difference in these angles with respect to the inner loop of the limaçon, and the circle.
Maple Solution
Initialize
Context Panel: Assign Name
r1=sinθ→assign
r2=1/5+cosθ→assign
Intersect the circle and limaçon
Write the equations as shown, and press the Enter key.
Context Panel: Solve≻Solve
r=r1,r=r2
r=sinθ,r=15+cosθ
→solve
r=45,θ=arctan43,r=−35,θ=arctan34−π
Point P in Figure 7.2.8(a) therefore has polar coordinates r,θ=−3/5,arctan3/4−π. The other solution r,θ=4/5,arctan4/3 corresponds to the intersection of the outer loop of the limaçon and the circle.
The inner loop of the limaçon passes through the origin when θ=arccos1/5−π, the solution of the equation r2=0 in the interval −π,0. This is most easily found with the Roots command in the Student Calculus1 package:
Access from the Context Panel
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Write the equation r2=0 and press the Enter key.
Context Panel: Student Calculus1≻Solve≻Find Roots See Roots dialog in Figure 7.2.8(b)
Figure 7.2.8(b) Roots dialog
r2=0
15+cosθ=0
→roots
−π+arccos15
Access directly
Student:-Calculus1:-Rootsr2=0,θ=−π..0 = −π+arccos15
If the circle is drawn with θ∈0,π, then it passes through point P when r1=3/5, so θ=arcsin3/5.
Calculation of areas A1 and A2 and their sum
12∫arctan34−πarccos15−π15+cosθ2 ⅆθ = −27100arctan34+27100arccos15−3506→assign to a nameA1
12∫0arcsin3/5sin2θ ⅆθ = −325+14arcsin35→assign to a nameA2
A1+A2 = −27100arctan34+27100arccos15−3506−325+14arcsin35→at 10 digits0.0899089628
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