Mathematics

Mathematics

Problem

Let r1=1+2sin⁡θr_1=1+2\sin\theta and r2=2cos⁡θr_2=2\cos\theta, traced for one full revolution 0≤θ≤2π0\le\theta\le 2\pi. A finite region is enclosed by alternating portions of these two curves. (a) Derive the equation that must be solved to locate all intersection angles of the curves on [0,2π][0,2\pi]. (b) Using either algebraic methods or numerical approximation, list all intersection angles in ascending order accurate to three decimal places. (c) Provide a reasoned argument—supported by a sign chart, inequalities, or a clearly labeled sketch—that identifies which curve is farther from the origin on each open interval determined in part (b). (d) Construct the full piece-wise integral expression for the total enclosed area and evaluate it to three decimal places. (e) One of your definite integrals may return a negative value before simplification. Explain why the final area reported in part (d) remains positive despite this sign issue.
Intuitive explanation

The key is recognizing that these two polar curves intersect at specific angles, and between those intersections, one curve lies outside the other. To find the enclosed region, you'll need to find where r1=r2r_1 = r_2, then use the polar area formula A=12∫r2 dθA = \frac{1}{2}\int r^2\,d\theta for each piece, integrating the outer curve minus the inner curve over the appropriate intervals. The "alternating portions" hint tells you the curves swap which is outer/inner at each intersection point.

Intuitive explanation

The key is recognizing that these two polar curves intersect at specific angles, and between those intersections, one curve lies outside the other. To find the enclosed region, you'll need to find where r1=r2r_1 = r_2, then use the polar area formula A=12∫r2 dθA = \frac{1}{2}\int r^2\,d\theta for each piece, integrating the outer curve minus the inner curve over the appropriate intervals. The "alternating portions" hint tells you the curves swap which is outer/inner at each intersection point.

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