Maths

As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jan08/Crazyint17.jpg

The astroid equation should probably read as follows:
astroid x=5cos(t)^3 ; y=5sin(t)^3;
where in the equation for y, the 5 is outside of the third power.

The area of an astroid is like a 4 cusped star, as shown in the Wiki-Encyclopedia illustration below:

圖片參考：http://i263.photobucket.com/albums/ii157/mathmate/math/Astroid_Wiki.png


The area is given as 3/8πa2 in the Wiki article:
http://en.wikipedia.org/wiki/Astroid
and the area for an n-sided astroid is given as
Area=(n-1)(n-2)πa2/n2
in the Mathworld article below:
http://mathworld.wolfram.com/Astroid.html.





The question appears to be an integration exercise and not just looking for the area by a formula, so the following gives the complete integration of the first quadrant of the astroid as an integration exercise. The complete area is therefore multiplied by four (4).



Given:

a=radius of astroid = 5
x=cos(t)^3
y=sin(t)^3
t=parameter (angle theta).
Differentiate the equation for x with respect to t,
dx=-3a cos(t)2sin(t) dt
y=a sin(t)3
y dx = 3a2cos(t)2sin(t)4 dt
Integrating by parts or by substitution between 0 and π/2, we obtain:
Area of first quadrant=(12*a^2*t-3*a^2*sin(4*t)-4*a^2*sin(2*t)^3)/64
which readily evaluates to 3πa2/32
Area of astroid=4*area of quadrant=3πa2/8