Chapter 7: Additional Applications of Integration
Section 7.2: Integration in Polar Coordinates
Working in polar coordinates, calculate the area common to the circle r1=sinθ and the inner loop of the limaçon r2=1/5+cosθ.
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
Figure 7.2.8(a) Intersection of circle and limaçon
The area of region R1 is given by the integral
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
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.
Context Panel: Assign Name
Intersect the circle and limaçon
Write the equations as shown, and press the Enter key.
Context Panel: Solve≻Solve
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
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
Student:-Calculus1:-Rootsr2=0,θ=−π..0 = −π+arccos⁡15
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 = −27100⁢arctan⁡34+27100⁢arccos⁡15−350⁢6→assign to a nameA1
12∫0arcsin3/5sin2θ ⅆθ = −325+14⁢arcsin⁡35→assign to a nameA2
A1+A2 = −27100⁢arctan⁡34+27100⁢arccos⁡15−350⁢6−325+14⁢arcsin⁡35→at 10 digits0.0899089628
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