Evaluate the line integral ydx+x2dy where C=C1+C2. C1 is the path of the straight line segment from the origin, (0,0) to the point (4,8). C2 is the path of the parabola y=−x2+5x+4 from the point (4,8) to the point (7,−10) .
parameterize for both the paths
??
so for c1 its x=4t y=8t and dx=4 dy=8?
yes
dx=4dt, dy=8dt
would the bounds from c2 be \[\int\limits_{8}^{-10}\] or \[\int\limits_{-10}^{8}\]
top one
i got -2940.667 as the final answer
but im wrong.
I don't understand why the bounds on c2 are 8 to -10 and not 4 to 7
so i should use the x instead of y? the example in my book uses y so i chose that.
each case is different, but here it seems that you would want to parameterize c2 as y(x) -> y(t)=y(x(t)) and x(t)=t
i have that the bounds just always confuse me... so it should be \[\int\limits_{4}^{7} -2t^3 + 4t^2 + 5t + 4\] ?
with the dts of course
yeah, that's what I get
i get -534... but its wrong.
I got a slightly different answer what did you get for the c1 integral?
176/3
c2 -592.66667
I was too afraid of an algebra error to do the c2 evaluation, so.... http://www.wolframalpha.com/input/?i=integral%20from%20t%3D4%20to%20t%3D7%20of%20-2t%5E3%2B4t%5E2%2B5t%2B4dt&t=crmtb01 you can see you likely made some small mistake
I agree with your c1
omg i didnt know you could do that!
uh-oh, don't get into cheating all the time :P your skills will turn to mush, I just do it to evade algebra stuff like this evaluation
thank you so much! i missed a couple classes on changing coordinates and this and the book wasnt helping me! but now im starting to get it!
happy to help, here's a good source imo: http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsIntro.aspx do we know if we got the last one right yet?
yep! so far they are all correct!
awesome :D
thanks. :)
@ganeshie8
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