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DCBaelar
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Homework Statement
Find the total area inside the circle r = 4 and below the line r=2csc[itex]\theta[/itex]
Homework Equations
[itex]\int[/itex][itex]^{b}_{a}[/itex] 1/2r[itex]^{2}[/itex][itex]\theta[/itex]d[itex]\theta[/itex]
The Attempt at a Solution
r=2/sin[itex]\theta[/itex][itex]\Rightarrow[/itex]rsin[itex]\theta[/itex]=2[itex]\Rightarrow[/itex]y=2
r=4[itex]\Rightarrow[/itex]=circle with radius 4 at center (0,0)
Point of Intersection:
4=2/sin[itex]\theta[/itex][itex]\Rightarrow4[/itex]sin[itex]\theta[/itex]=2[itex]\Rightarrow[/itex]sin[itex]\theta[/itex]=2/4[itex]\Rightarrow[/itex][itex]\theta[/itex]=[itex]\pi[/itex]/6
[itex]\int[/itex][itex]^{pi/6}_{0}[/itex] 1/2 (4-2csc[itex]\theta[/itex])[itex]^{2}[/itex]d[itex]\theta[/itex]
[itex]\int[/itex][itex]^{pi/6}_{0}[/itex] 1/2 (16-4csc[itex]^{2}[/itex][itex]\theta[/itex]d[itex]\theta[/itex])
[itex]\int[/itex][itex]^{pi/6}_{0}[/itex] (8-2csc[itex]^{2}[/itex][itex]\theta[/itex]d[itex]\theta[/itex])
(8[itex]\theta[/itex]+2cot[itex]\theta[/itex])|[itex]^{pi/6}_{0}[/itex]
(8*pi/6)+2cot(pi/6)-(8*0)-2cot(0)
4pi/3+2√3-0-undefined
And that's my problem..the undefined 2cot.
I think where I went wrong is that sinθ=1/2 at [itex]\pi[/itex]/6 and 5[itex]\pi[/itex]/6 and thus my boundaries of integration should be [itex]\pi[/itex]/6 and 5[itex]\pi[/itex]/6.
Am I on the right track?
Thanks for any feedback/assistance.
Jason
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