area


by ILoveBaseball
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ILoveBaseball
ILoveBaseball is offline
#1
Apr7-05, 05:15 AM
P: 30
Find the area of the region bounded by: [tex]r= 6-2sin(\theta)[/tex]

here's what i did:

[tex] 6-2sin(\theta) = 0[/tex]
[tex] sin(\theta) = 1/3[/tex]

so the bounds are from arcsin(-1/3) to arcsin(1/3) right?

my integral:
[tex]\int_{-.339}^{.339} 1/2*(6-2sin(\theta))^2[/tex]

i get a answer of 0.6851040673*10^11, and it's wrong. all my steps seems to be correct, i cant figure out the problem.
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marlon
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#2
Apr7-05, 05:36 AM
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What is the answer that you should have got ???

marlon
asrodan
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#3
Apr7-05, 09:46 PM
P: 40
[tex]r= 2sin(\theta)[/tex] is an ellipse so [tex]r= 6-2sin(\theta)[/tex] is just shifting and stretching it.

Therefore the bounds on [tex] \theta [/tex] are [tex] 0 \leq \theta \leq 2 \pi [/tex]

dextercioby
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#4
Apr7-05, 09:55 PM
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P: 11,863

area


I agree.It's a shifted & stretched ellipse.Pay attention with the numbers...You can't get a big value for the area.It's ~100...

Daniel.

[itex] 38\pi [/itex] to be exact.
ILoveBaseball
ILoveBaseball is offline
#5
Apr7-05, 10:35 PM
P: 30
Quote Quote by asrodan
[tex]r= 2sin(\theta)[/tex] is an ellipse so [tex]r= 6-2sin(\theta)[/tex] is just shifting and stretching it.

Therefore the bounds on [tex] \theta [/tex] are [tex] 0 \leq \theta \leq 2 \pi [/tex]
can you explain it to me agian? i dont really understand it that well. are you saying if it's an ellipse, the bounds will always be from 0 ->2pi?
dextercioby
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#6
Apr7-05, 10:50 PM
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Yes,it's like for a circle,or for any closed curve enclosing the origin inside it...

Daniel.
ILoveBaseball
ILoveBaseball is offline
#7
Apr7-05, 11:49 PM
P: 30
ah, i get it now. thank you


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