Chapter 5: Double Integration
Section 5.7: Double Integration in Polar Coordinates
Calculate the area that is inside both the rose r=sin2 θ and the circle r=sinθ.
Figure 5.7.10(a) shows the rose and circle, and, in green and gray, the region whose area is to be calculated.
Figure 5.7.10(b) is an animation in which the two bounding 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 5.7.10(b), Figure 5.7.10(c) is a graph of r=sin2 θ (in black) and r=sinθ (in red). It suggests that the curves first intersect for θ=±π/3. This would allow the area of the green region in the first quadrant to be calculated. This region is separated from the gray region by the line y=x3. The first-quadrant loop of the rose that bounds this gray region is traced for θ∈π/3,π/2. The first-quadrant area is the sum of the green and gray areas, and the total area to be computed is twice this, by symmetry.
use plots in
Figure 5.7.10(a) Region
Figure 5.7.10(b) Animation
Figure 5.7.10(c) Intersections
The area of the region shaded in green and gray in Figure 5.7.10(a) is
2∫02π/3∫0sinθr ⅆr ⅆθ+2∫2π/3π/2∫0sin2 θr ⅆr ⅆθ = π/4−33/16 ≐ 0.461
Maple Solution - Interactive
The task template in Tables 5.7.10(a) and 5.7.10(b) can be used to visualize the region of integration over which a given iterated integral acts. Table 5.7.10(a) calculates the area in the first-quadrant green region in Figure 5.7.10(a); the second, the first-quadrant gray area. In each case, 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.
Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Polar
Evaluate ∬RΨr,θ dA and Graph R
Area Element dA
r dr dθ
r dθ dr
Value of Integral
Table 5.7.10(a) Task template for visualizing integration over upper half of the limaçon
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.
Table 5.7.10(b) Task template for visualizing integration over inner loop of the limaçon
Combining the results in Tables 5.6.10(a - b) gives the requisite area as
2π12−316+π24− 332 = 23 π24−3332 = π4−3316 ≐ 0.461
The details of an interactive calculation of the required area appear in Table 5.7.10(c).
Tools≻Load Package: Student Calculus 1
Find the zeros of sinθ=sin2 θ
Write the equation.
Student Calculus1≻Solve≻Find Roots
Complete the dialog as per Figure 5.7.10(d)
Figure 5.7.10(d) Roots dialog
Implement and evaluate the iterated integration
Calculus palette: Iterated double integral template
Context Panel: Evaluate and Display Inline
2∫02π/3∫0sinθr ⅆr ⅆθ+∫2π/3π/2∫0sin2 θr ⅆr ⅆθ = 316⁢3+14⁢π
Table 5.7.10(c) Details of the interactive calculation of the required area.
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