Question

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.

Answer 1

did you want to check an answer, or are you stuck?

Answer 2

\[512\int\limits_{0}^{\pi} \cos^2t sint - \sin^3t\] im stuck at how to integrate this.

Answer 3

use\[u=\cos t\]and\[\sin^2t=1-\cos^2t\]

Answer 4

so du =-sinx

Answer 5

how did you get cos^2t*sint for x^2+y^2dx ?

Answer 6

i keep getting confused but heres where i started \[\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(-8sint) + 2(8cost)(8sint)(8cost)dt\]

Answer 7

yes, and that can be written\[\int_0^\pi512(\sin^2t+\cos^2t)(-\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt\]

Answer 8

\[-512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint\] which becomes this

Answer 9

yes, which you can recombine to form this\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]

Answer 10

so would you use u=cost and du=-sint ?

Answer 11

yep

Answer 12

or could you do \[sint(\cos^2t-1)\] which becomes \[\sin^3t\] ?

Answer 13

and how to you propose to integrate that?

Answer 14

oh yeah.haha.

Answer 15

\[-512\int\limits_{1}^{-1}- 2u^2du - du \] ?

Answer 16

i mean + du

Answer 17

yeah, looks good

Answer 18

oh wait, how did 512 get negative?

Answer 19

im confused because du=-sint and the sint is positive so (-1/-1) would need to be multiplied right?

Answer 20

yeah, but you already covered that by changing thethe signs og the terms

Answer 21

however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it

Answer 22

\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt\]

Answer 23

sorry I messed up the bounds on the last integral, they should be 0 to pi

Answer 24

and the answer is 1706.66666666667?

Answer 25

nevermind thats wrong.

Answer 26

\[512\int_0^\pi2\cos^2t\sin t-\sin tdt\]\[u=\cos t\implies du=-\sin tdt\]\[512\int_{1}^{-1}- 2u^2du +\int_0^{\pi} -\sin tdt\]let me see what I get...

Answer 27

\[512\left(\left.-\frac23u^3\right|_1^{-1}+\left.\cos t\right|_0^\pi\right)\]\[512(\frac43-2)=512(-\frac23)\]but I always could have made a mistake...

Answer 28

its right! yay!

Answer 29

sweet, I just hope you find your (likely algebra-based) mistake

Answer 30

thats what i got to too! thanks so much!

Answer 31

welcome!