http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 28 Jul 2014 14:28:14 -0500http://calc3.askbot.com/question/186/how-do-i-get-the-bounds-of-integration-for-line-integrals/I am currently working on the homework for 16.2. Number one, for example, states: **"Compute $\int\limits_Cxy^2ds$ along the line segment from (1,2,0) to (2,1,3)"**. I am setting it up as follows: For my parameterized line I get: $$\vec{p}(t) = \langle1,2,0\rangle + t\langle1,-1,3\rangle$$ Giving me: $$x=t+1$$ $$ x' = 1$$ $$y=2-t$$ $$ y'= -1$$ $$z=3t$$ $$ z'= 3$$ Then setting up an integral to compute: $$\int(t+1)(2-t)^2\sqrt{(1)^2+(-1)^2+(3)^2}dt$$ What I am not sure of is how to know what my bounds of integration are. Do I plug my point values in for x, y, and z? Or in this case, for x and y and then solve for t? Or just for x? I would love any help on this or to know if what I have done so far is correct or not correct...Thanks! Mon, 28 Jul 2014 12:06:32 -0500http://calc3.askbot.com/question/186/how-do-i-get-the-bounds-of-integration-for-line-integrals/http://calc3.askbot.com/question/186/how-do-i-get-the-bounds-of-integration-for-line-integrals/?answer=187#post-id-187Looks good so far! Or at least, this is what I have, and when I compute the integral I get the answer the book has. As for the bounds of integration, I believe it is from $0\leq t \leq 1$ due to our parametrization of the line $\vec{p}(t) = \langle1,2,0\rangle + t\langle1,-1,3\rangle$. If I'm not mistaken, whenever we parametrize a line segment, $0\leq t \leq 1$ is our bounds. When we get into parametrizing different shapes such as the top half of the unit circle in problem 4, our bounds change. For problem 4, I believe, our bounds would be $0\leq t\leq \pi$. I hope this helps, but if anyone else can add to this that'd be great.Mon, 28 Jul 2014 14:28:14 -0500http://calc3.askbot.com/question/186/how-do-i-get-the-bounds-of-integration-for-line-integrals/?answer=187#post-id-187