Area enclosed by the astroid x^(2/3)/9 + y^(2/3)/4 = 1

[mmmboh]

Find the area of x^2/3/9 +y^2/3/4 =1
The parametrizations I used are x=27cos³t y=8sin³t

Area = $$\int xdy = \int_{0}^{2\pi} 27cos^3(t)24sin^2(t)cos(t)= \int_{0}^{2\pi} 27cos^4(t)24sin^2(t)=648((2\pi)/16+sin(4\pi)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi$$
Is this right?

 

[HallsofIvy]

Your formula is correct but the result is not at all what I get. Please show how you did the integral.

 

[mmmboh]

$$\int xdy = \int_{0}^{2\pi} 27cos^3(t)24sin^2(t)cos(t)= \int_{0}^{2\pi} 27cos^4(t)24sin^2(t)=648((2\pi)/16+sin(4\pi)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi$$

 

[mmmboh]

Can someone help please?

 

[mathman44]

I get 81pi too. Can anyone confirm this?

 

[mathman44]

Anyone, please?

 

[Dick]

Yes. 81*pi.