The area bounded by the curve x^(2/3)+y^(2/3)=a^(2/3) is given by 4*integral(ydx) between limits [0,a]. These limits are defined by substituting y=0 in the curve equation: x^(2/3)=a^(2/3), (x^(1/3))^2=a^(2/3), x^(1/3)=+a^(1/3), so x=+a. y=(a^(2/3)-x^(2/3))^(3/2), therefore the integral becomes ydx=(a^(2/3)-x^(2/3))^(3/2)dx [0,a]. The curve is split into 4 quadrants because of the symmetry of the astroid about both axes, so there is only a need to find the area of one quadrant, which explains the reason for the limits between 0 and a, and for multiplying the integral by 4.
Let x=a(sinz)^3, so dx=3a(sinz)^2coszdz. Integral(a^2/3-x^2/3))^(3/2)dx) [0,a] becomes:
a*integral((1-(x/a)^(2/3))^(3/2)dx) [0,a]=
3a^2*integral(1-(sinz)^2)^(3/2)(sinz)^2coszdz) [0,(pi)/2]=
3a^2*integral((cosz)^3(sinz)^2coszdz)=3a^2integral((cosz)^4(sinz)^2dz) [0,(pi)/2].
Consider the following trig identities.
cos4z=2(cos2z)^2-1; 4(cos2z)^2=2cos4z+2 and cos2z=2(cosz)^2-1,
so 2(cosz)^2=cos2z+1 and 8(cosz)^2=4cos2z+4
cos4z=2(2(cosz)^2-1)^2-1=2(4(cosz)^4-4(cosz)^2+1)-1=8(cosz)^4-8(cosz)^2+1.
Therefore, (cosz)^4=(cos4z+8(cosz)^2-1)/8=(cos4z+4cos2z+3)/8.
cos2z=1-2(sinz)^2, so (sinz)^2=(1-cos2z)/2.
Therefore, 3a^2integral((cosz)^4(sinz)^2dz)=
(3a^2/16)integral((cos4z+4cos2z+3)(1-cos2z)dz) [0,(pi)/2]=
(3a^2/16)integral((cos4z+4cos2z+3-cos4zcos2z-4(cos2z)^2-3cos2z)dz) [0,(pi)/2]=
(3a^2/16)integral((cos4z+4cos2z+3-cos4zcos2z-2cos4z-2-3cos2z)dz) [0,(pi)/2]=
(3a^2/16)integral((-cos4z+cos2z+1-cos4zcos2z)dz) [0,(pi)/2].
Now consider the integral of -cos4zcos2zdz. Let dv=cos2zdz and u=cos4z,
then v=sin2z/2 and du=-4sin4zdz.
Integrating by parts: integral(-cos4zcos2zdz)=-cos4zsin2z/2+integral(2sin2zsin4zdz)=
-cos4zsin2z/2+integral(4(sin2z)^2cos2zdz) because sin4z=2sin2zcos2z.
Let p=(sin2z)^2 then 2sin2z=2p^(1/2) and dp=2sin2zcos2zdz=2p^(1/2).cos2zdz, so
cos2zdz=(p^-(1/2)/2)dp and integral(4(sin2z)^2cos2zdz)=
integral((2p^(1/2)dp)=(4/3)p^(3/2)=(4/3)(sin2z)^3=integral(4(sin2z)^2cos2zdz)
Integral(-cos4zcos2zdz)=-cos4zsin2z/2+(4/3)(sin2z)^3
We can now return to the original integral:
(3a^2/16)integral((-cos4z+cos2z+1-cos4zcos2z)dz) [0,(pi)/2]=
(3a^2/16)(-sin4z/4+sin2z/2+z-cos4zsin2z/2+(4/3)(sin2z)^3) [0,(pi)/2].
All the terns involving sin2z or sin4z become zero under the limits for z=0 or (pi)/2, and all that remains is 3a^2(pi)/32. This is the quadrant area, so the total area is 4 times this=3a^2(pi)/8.