• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter mmmboh
  • Start date
  • #1
407
0
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?
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,793
922
Your formula is correct but the result is not at all what I get. Please show how you did the integral.
 
  • #3
407
0
[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]
 
Last edited:
  • #4
407
0
Can someone help please?
 
  • #5
206
0
I get 81pi too. Can anyone confirm this?
 
Last edited:
  • #6
206
0
Anyone, please?
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
618
Yes. 81*pi.
 

Related Threads for: Area of Astroid? Or something like it. Is this right?

  • Last Post
Replies
3
Views
9K
Replies
5
Views
18K
Replies
7
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
16
Views
414
Top