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

Find the area of x2/3/9 +y2/3/4 =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?

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

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

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I get 81pi too. Can anyone confirm this?

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