Question

45–60. Areas of regions Find the area of the following regions.The region inside the curve r = √(cos θ) and inside the circle r = 1/√2 in the first quadrant

Accepted Answer

Identify the curves given: the polar curve is $$ r = \sqrt{\cos 	heta} $$ and the circle is $$ r = \frac{1}{\sqrt{2}} . We $$are interested in the region inside both curves in the first quadrant, where $$ 	heta $$ ranges from 0 to $$ \frac{\pi}{2} . $$Find the points of intersection between the two curves by setting $$ \sqrt{\cos 	heta} = \frac{1}{\sqrt{2}} . $$Square both sides to get $$ \cos 	heta = \frac{1}{2} $$, then solve for $$ 	heta  in $$the first quadrant. Determine the limits of integration for $$ 	heta $$ based on the intersection point and the first quadrant. The region inside both curves will be bounded by $$ 	heta = 0 $$ and the intersection angle found in the previous step. Set up the integral for the area of the region inside both curves. Since the region is inside both curves, the radius at each $$ 	heta  is $$the smaller of the two radii. This means the area is the integral from 0 to the intersection angle of the area under $$ r = \sqrt{\cos 	heta} $$, plus the integral from the intersection angle to $$ \frac{\pi}{2}  of $$the area under $$ r = \frac{1}{\sqrt{2}} . $$Use the polar area formula: $$ A = \frac{1}{2} \int_{\alpha}^{\beta} r(	heta)^2 \, d	heta . $$Write the final expression for the area as the sum of two integrals: $$ A = \frac{1}{2} \int_0^{	heta_0} \cos 	heta \, d	heta + \frac{1}{2} \int_{	heta_0}^{\frac{\pi}{2}} \frac{1}{2} \, d	heta $$, where $$ 	heta_0  is $$the intersection angle found earlier.

45–60. Areas of regions Find the area of the following regions.

The region inside the curve r = √(cos θ) and inside the circle r = 1/√2 in the first quadrant

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Key Concepts

Polar Coordinates and Curves

Area Calculation in Polar Coordinates

Determining Intersection Points and Integration Limits