The total area of the astroid is

4 times the first quadrant area

2 times the first quadrant area

8 times the first quadrant area

equal to the first quadrant area

Area = 4 integral y dx becomes

12a^2 integral sin^4 t cos^2 t dt over 0 to π/2

12a^2 integral sin^2 t cos^4 t dt

4a^2 integral sin^3 t cos^3 t dt

Short answer

Use x = a cos³t, y = a sin³t; area = 4∫₀^a y dx = 12a² ∫₀^{π/2} sin⁴t cos²t dt. Beta function yields π/32, giving area = 3πa²/8.

Show that the area of the curve x^(2/3) + y^(2/3) = a^(2/3) is (3/8)πa².

Astroid: x^(2/3) + y^(2/3) = a^(2/3). It is symmetrical about both axes, so the total area is four times the area in the first quadrant.

Use x = a cos³t, y = a sin³t. Then dx = −3a cos²t sin t dt. As x goes from 0 to a, t goes from π/2 to 0.

Area = 4 ∫₀^a y dx
     = 4 ∫_(π/2)^0 a sin³t (−3a cos²t sin t) dt
     = 12a² ∫₀^(π/2) sin⁴t cos²t dt

Beta function: ∫₀^(π/2) sin⁴t cos²t dt = Γ(5/2)Γ(3/2)/(2Γ(4)).

Γ(5/2) = 3√π/4, Γ(3/2) = √π/2, Γ(4) = 6.

= (3√π/4)(√π/2) / 12 = (3π/8)/12 = π/32.

Area = 12a² · π/32

Area = 3πa²/8. Shown.

✓ Answer checked by Claude Opus 5.5 (high effort)