Answer to Question #176653 in Calculus for Joshua

Question

Find the area of the curve r^2=3-4cos^2(0), symmetrical to origin and Ox.

Answer 1

Crazy condition - clearly in the original it was about the area of ​​the figure bounded by the curve. Fortunately, it (the figure) contains the origin, so the method is known - 2∆S = r² (α) ∆α (if the angular argument is α) and integrate for health. The devil is in the details, here - in the area of ​​changing the angular argument. It makes no sense to give the entire area, it is enough to indicate the segment [π / 6; π / 2] in the first quadrant, because due to symmetry, the entire area is 4 times larger than the part of the figure that is located in the 1st quadrant.

$S=(\frac 4 2)\int^{\frac \pi 2}_{\frac \pi 6}[3-4cos² (α)]dα= 2\int^{\frac \pi 2}_{\frac \pi 6}[1-cos (2α)]dα= 2\frac \pi 3+sin(\frac \pi 3)=2\frac \pi 3+\frac {\sqrt{3}} 2$