Area of Astroid? Or something like it. Is this right?

mmmboh

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

Area = [tex]\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[/tex]



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

[tex]\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[/tex]

mmmboh

Can someone help please?

mathman44

I get 81pi too. Can anyone confirm this?

mathman44

Anyone, please?

Dick

Yes. 81*pi.