x ( θ ) = a ( θ + s i n ( θ ) ) y ( θ ) = a ( 1 + s i n ( θ ) ) x(\theta)=a(\theta+sin(\theta))\\
y(\theta)=a(1+sin(\theta))\\ x ( θ ) = a ( θ + s in ( θ )) y ( θ ) = a ( 1 + s in ( θ ))
area calculation formula:
P = 2 π ∫ θ 1 θ 2 y ( θ ) ∗ ( x ′ ( θ ) ) 2 + ( y ′ ( θ ) ) 2 d θ x ′ ( θ ) = a + a ∗ c o s ( θ ) y ′ ( θ ) = − a s i n ( θ ) P=2\pi \displaystyle\intop _{\theta_1} ^{\theta_2} y(\theta)*\sqrt{(x'(\theta))^2+(y'(\theta))^2}d\theta\\
x'(\theta)=a+a*cos(\theta)\\
y'(\theta)=-asin(\theta)\\ P = 2 π θ 1 ∫ θ 2 y ( θ ) ∗ ( x ′ ( θ ) ) 2 + ( y ′ ( θ ) ) 2 d θ x ′ ( θ ) = a + a ∗ cos ( θ ) y ′ ( θ ) = − a s in ( θ )
after substitution and simplification
P = 2 π ∫ 0 π a ∗ ( 1 + c o s ( θ ) ) ∗ a ∗ 2 ∗ c o s ( θ / 2 ) d θ = = 4 π a 2 ∫ 0 π ( c o s ( θ / 2 ) + c o s ( θ ) ∗ c o s ( θ / 2 ) ) d θ = = 32 π a 2 / 3 P=2\pi\displaystyle\intop _{0} ^{\pi} a*(1+cos(\theta))*a*2*cos(\theta/2)d\theta=\\
=4\pi a^2\displaystyle\intop_{0}^{\pi}(cos(\theta/2)+cos(\theta)*cos(\theta/2))d\theta=\\
=32\pi a^2/3 P = 2 π 0 ∫ π a ∗ ( 1 + cos ( θ )) ∗ a ∗ 2 ∗ cos ( θ /2 ) d θ = = 4 π a 2 0 ∫ π ( cos ( θ /2 ) + cos ( θ ) ∗ cos ( θ /2 )) d θ = = 32 π a 2 /3
since we calculated only half of the figure, the result should be multiplied by 2
answer: 64 π ∗ a 2 / 3 64 \pi *a^2/3 64 π ∗ a 2 /3
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