Question

Find the area of the curve r^2 = 3-4cos^2∅, symmetrical to origin and Ox.

Answer 1

Given, the curve r² = 3-4cos²∅, symmetrical to origin.

Area of region above the origin=

$=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} r^2 d\varnothing\newline =\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 3-4cos^2\varnothing d\varnothing\newline =\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 3-4\frac{1-cos2\varnothing}{2} d\varnothing\newline =\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 1+cos2\varnothing d\varnothing\newline =\frac{1}{2}[ \varnothing + \frac{sin2\varnothing}{2} ]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\newline =\frac{1}{2}[ \frac{5\pi}{6} -\frac{\sqrt3}{4} -{\frac{\pi}{6}}-\frac{\sqrt3}{4}]\newline =\frac{1}{2}[ \frac{2\pi}{3} -\frac{\sqrt3}{2}]\newline =0.614 \text{square units}$

Total area=2×0.614=1.228 square units.