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Areaby ILoveBaseball
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#1
Apr705, 05:15 AM

P: 30

Find the area of the region bounded by: [tex]r= 62sin(\theta)[/tex]
here's what i did: [tex] 62sin(\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*(62sin(\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. 


#3
Apr705, 09:46 PM

P: 40

[tex]r= 2sin(\theta)[/tex] is an ellipse so [tex]r= 62sin(\theta)[/tex] is just shifting and stretching it.
Therefore the bounds on [tex] \theta [/tex] are [tex] 0 \leq \theta \leq 2 \pi [/tex] 


#4
Apr705, 09:55 PM

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P: 11,893

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. 


#5
Apr705, 10:35 PM

P: 30




#6
Apr705, 10:50 PM

Sci Advisor
HW Helper
P: 11,893

Yes,it's like for a circle,or for any closed curve enclosing the origin inside it...
Daniel. 


#7
Apr705, 11:49 PM

P: 30

ah, i get it now. thank you



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