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Evaluate the line integral
(x2+y2)dx+2xydy
where is the path of the semicircular arc of the circle x2+y2=64 starting at (8,0) and ending at (−8,0) going counterclockwise.
 one year ago
 one year ago
Evaluate the line integral (x2+y2)dx+2xydy where is the path of the semicircular arc of the circle x2+y2=64 starting at (8,0) and ending at (−8,0) going counterclockwise.
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.1
did you want to check an answer, or are you stuck?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
\[512\int\limits_{0}^{\pi} \cos^2t sint  \sin^3t\] im stuck at how to integrate this.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
use\[u=\cos t\]and\[\sin^2t=1\cos^2t\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
how did you get cos^2t*sint for x^2+y^2dx ?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
i keep getting confused but heres where i started \[\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(8sint) + 2(8cost)(8sint)(8cost)dt\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, and that can be written\[\int_0^\pi512(\sin^2t+\cos^2t)(\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt\]
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
\[512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint\] which becomes this
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, which you can recombine to form this\[512\int_0^\pi2\cos^2t\sin t\sin tdt\]
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
so would you use u=cost and du=sint ?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
or could you do \[sint(\cos^2t1)\] which becomes \[\sin^3t\] ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and how to you propose to integrate that?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
\[512\int\limits_{1}^{1} 2u^2du  du \] ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh wait, how did 512 get negative?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
im confused because du=sint and the sint is positive so (1/1) would need to be multiplied right?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, but you already covered that by changing thethe signs og the terms
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[512\int_0^\pi2\cos^2t\sin t\sin tdt\]\[u=\cos t\implies du=\sin tdt\]\[512\int_{1}^{1} 2u^2du +\int_1^{1} \sin tdt\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
sorry I messed up the bounds on the last integral, they should be 0 to pi
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
and the answer is 1706.66666666667?
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
nevermind thats wrong.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[512\int_0^\pi2\cos^2t\sin t\sin tdt\]\[u=\cos t\implies du=\sin tdt\]\[512\int_{1}^{1} 2u^2du +\int_0^{\pi} \sin tdt\]let me see what I get...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[512\left(\left.\frac23u^3\right_1^{1}+\left.\cos t\right_0^\pi\right)\]\[512(\frac432)=512(\frac23)\]but I always could have made a mistake...
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
its right! yay!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
sweet, I just hope you find your (likely algebrabased) mistake
 one year ago

meggiesmermaidBest ResponseYou've already chosen the best response.0
thats what i got to too! thanks so much!
 one year ago
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