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Find the area of the region that lies inside of the curve r = 3cos(theta) and outside of the curve r = 1 + cos(theta). Be sure to sketch the graph of the enclosed area as well.

User Mattexx
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Final answer:

To find the area between the two polar curves r = 3cos(θ) and r = 1 + cos(θ), intersecting points are first determined by solving 3cos(θ) = 1 + cos(θ). The enclosed area is then calculated by integrating the difference of the squares of the polar functions with respect to θ, using the found intersection points as limits.

Step-by-step explanation:

The question involves finding the area of the region that is enclosed by two polar curves: r = 3cos(θ) and r = 1 + cos(θ). To find this area, one must first sketch the graph of these polar equations to visualize the region being referred to. The area can then be found using the formula for the area of a sector in polar coordinates, A = 1/2 ∫(r^2)dθ, where the limits of integration are the angles where the two curves intersect.

First, the intersections of the curves must be found by setting the two equations equal to each other and solving for θ:
3cos(θ) = 1 + cos(θ). Simplifying, this gives cos(θ) = 1/2, which implies that θ = π/3 and 5π/3 (since cosine is positive in the first and fourth quadrants). These will be the limits of integration.

The area enclosed by the curves is the area under the curve r = 3cos(θ) minus the area under the curve r = 1 + cos(θ) over the range from θ = π/3 to θ = 5π/3. So, the integral to find the area would be A = 1/2 ∫π/35π/3((3cos(θ))^2 - (1 + cos(θ))^2)dθ.

By calculating this integral, we can find the area of the enclosed region.

User Nehemie KOFFI
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