anonymous
  • anonymous
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.
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

TuringTest
  • TuringTest
did you want to check an answer, or are you stuck?
anonymous
  • anonymous
\[512\int\limits_{0}^{\pi} \cos^2t sint - \sin^3t\] im stuck at how to integrate this.
TuringTest
  • TuringTest
use\[u=\cos t\]and\[\sin^2t=1-\cos^2t\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so du =-sinx
TuringTest
  • TuringTest
how did you get cos^2t*sint for x^2+y^2dx ?
anonymous
  • anonymous
i keep getting confused but heres where i started \[\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(-8sint) + 2(8cost)(8sint)(8cost)dt\]
TuringTest
  • TuringTest
yes, and that can be written\[\int_0^\pi512(\sin^2t+\cos^2t)(-\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt\]
anonymous
  • anonymous
\[-512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint\] which becomes this
TuringTest
  • TuringTest
yes, which you can recombine to form this\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]
anonymous
  • anonymous
so would you use u=cost and du=-sint ?
TuringTest
  • TuringTest
yep
anonymous
  • anonymous
or could you do \[sint(\cos^2t-1)\] which becomes \[\sin^3t\] ?
TuringTest
  • TuringTest
and how to you propose to integrate that?
anonymous
  • anonymous
oh yeah.haha.
anonymous
  • anonymous
\[-512\int\limits_{1}^{-1}- 2u^2du - du \] ?
anonymous
  • anonymous
i mean + du
TuringTest
  • TuringTest
yeah, looks good
TuringTest
  • TuringTest
oh wait, how did 512 get negative?
anonymous
  • anonymous
im confused because du=-sint and the sint is positive so (-1/-1) would need to be multiplied right?
TuringTest
  • TuringTest
yeah, but you already covered that by changing thethe signs og the terms
TuringTest
  • TuringTest
however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it
TuringTest
  • TuringTest
\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt\]
TuringTest
  • TuringTest
sorry I messed up the bounds on the last integral, they should be 0 to pi
anonymous
  • anonymous
and the answer is 1706.66666666667?
anonymous
  • anonymous
nevermind thats wrong.
TuringTest
  • TuringTest
\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_0^{\pi} -\sin tdt\]let me see what I get...
TuringTest
  • TuringTest
\[512\left(\left.-\frac23u^3\right|_1^{-1}+\left.\cos t\right|_0^\pi\right)\]\[512(\frac43-2)=512(-\frac23)\]but I always could have made a mistake...
anonymous
  • anonymous
its right! yay!
TuringTest
  • TuringTest
sweet, I just hope you find your (likely algebra-based) mistake
anonymous
  • anonymous
thats what i got to too! thanks so much!
TuringTest
  • TuringTest
welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.