Area of Astroid? Or something like it. Is this right?by mmmboh Tags: area, astroid, calculus, parametrize 

#1
Mar210, 12:00 AM

P: 408

Find the area of x^{2/3}/9 +y^{2/3}/4 =1
The parametrizations I used are x=27cos^{3}t y=8sin^{3}t 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)/64sin(8\pi)/64sin(12\pi)/192) = 81\pi[/tex] Is this right? 



#2
Mar210, 09:10 AM

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.




#3
Mar210, 10:33 AM

P: 408

[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)/64sin(8\pi)/64sin(12\pi)/192) = 81\pi[/tex]




#4
Mar310, 02:47 PM

P: 408

Area of Astroid? Or something like it. Is this right?
Can someone help please?




#5
Mar510, 02:45 PM

P: 210

I get 81pi too. Can anyone confirm this?




#6
Mar610, 01:05 PM

P: 210

Anyone, please?




#7
Mar610, 02:50 PM

Sci Advisor
HW Helper
Thanks
P: 25,167

Yes. 81*pi.



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