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

by mmmboh
Tags: area, astroid, calculus, parametrize
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mmmboh
#1
Mar2-10, 12:00 AM
P: 408
Find the area of x2/3/9 +y2/3/4 =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?
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HallsofIvy
#2
Mar2-10, 09:10 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363
Your formula is correct but the result is not at all what I get. Please show how you did the integral.
mmmboh
#3
Mar2-10, 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)/64-sin(8\pi)/64-sin(12\pi)/192) = 81\pi[/tex]

mmmboh
#4
Mar3-10, 02:47 PM
P: 408
Area of Astroid? Or something like it. Is this right?

Can someone help please?
mathman44
#5
Mar5-10, 02:45 PM
P: 210
I get 81pi too. Can anyone confirm this?
mathman44
#6
Mar6-10, 01:05 PM
P: 210
Anyone, please?
Dick
#7
Mar6-10, 02:50 PM
Sci Advisor
HW Helper
Thanks
P: 25,242
Yes. 81*pi.


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