enter image description here

I tried following the example in my book, and I got stuck when trying to take the integral for the second part with $y = x^2$. I'm not completely confident on what to parameterize with here. I tried replacing $x^2$ with $y$, then in the radical I had $4 + y^3$, but when trying to integrate that with u substitution I couldn't finish it. I'm not sure I'm setting everything up right.

I'm also not sure how to tell which integral goes with which segment. Do you just assume the first goes with the first and second with the second?



the first segment can be parametrised as

$$x=0+2t \;\;\;,\;\; y=0+t $$

the integral along this segment is

$$\int_0^1 \Bigl ((2t+2t)(2dt)+4t^2 (dt)\Bigr)=$$

$$4\int_0^1t (2+t)dt=\frac {16}{3}$$

You can do it for the second segment defined by

$$x=2+t \;\;\;, \;\; y=1-t$$

  • $\begingroup$ also can you explain how you parametrized those equations? $\endgroup$ – 2316354654 Nov 26 '17 at 20:00
  • $\begingroup$ @2316354654 the line AB is parametrized by $x=x_A+t (x_B-x_A) $ and $y=... $ $\endgroup$ – hamam_Abdallah Nov 26 '17 at 20:14
  • $\begingroup$ why is your integral from 1 to 0? I thought it would be 2 to 0. $\endgroup$ – 2316354654 Nov 26 '17 at 21:02
  • $\begingroup$ is the integral always from 1 to 0when you parameterize in terms of t? $\endgroup$ – 2316354654 Nov 26 '17 at 21:15
  • 1
    $\begingroup$ @2316354654 Yes. t=0 gives the first point of the segment and t=1 gives the second. $\endgroup$ – hamam_Abdallah Nov 27 '17 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.