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meggiesmermaid
 3 years 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.
meggiesmermaid
 3 years 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.

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1did you want to check an answer, or are you stuck?

meggiesmermaid
 3 years ago
Best 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.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1use\[u=\cos t\]and\[\sin^2t=1\cos^2t\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1how did you get cos^2t*sint for x^2+y^2dx ?

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

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

meggiesmermaid
 3 years ago
Best 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

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

meggiesmermaid
 3 years ago
Best ResponseYou've already chosen the best response.0so would you use u=cost and du=sint ?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1and how to you propose to integrate that?

meggiesmermaid
 3 years ago
Best ResponseYou've already chosen the best response.0\[512\int\limits_{1}^{1} 2u^2du  du \] ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1oh wait, how did 512 get negative?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, but you already covered that by changing thethe signs og the terms

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1sorry I messed up the bounds on the last integral, they should be 0 to pi

meggiesmermaid
 3 years ago
Best ResponseYou've already chosen the best response.0and the answer is 1706.66666666667?

meggiesmermaid
 3 years ago
Best ResponseYou've already chosen the best response.0nevermind thats wrong.

TuringTest
 3 years ago
Best 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...

TuringTest
 3 years ago
Best 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...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1sweet, I just hope you find your (likely algebrabased) mistake

meggiesmermaid
 3 years ago
Best ResponseYou've already chosen the best response.0thats what i got to too! thanks so much!
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