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

by mmmboh
Tags: area, astroid, calculus, parametrize
 P: 408 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,879 Your formula is correct but the result is not at all what I get. Please show how you did the integral.
 P: 408 $$\int xdy = \int_{0}^{2\pi} 27cos^3(t)24sin^2(t)cos(t)= \int_{0}^{2\pi} 27cos^4(t)24sin^2(t)=648((2\pi)/16+sin(4\pi)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi$$
P: 408

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

Can someone help please?
 P: 210 I get 81pi too. Can anyone confirm this?
 P: 210 Anyone, please?
 Sci Advisor HW Helper Thanks P: 25,167 Yes. 81*pi.

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