Cal help

[junhoma]

I have to parametraze piecewise smooth curve for line integral.

One piece goes from points(2, 1) to points ( 2, 4) and another piece goes from points (2, 4) to points ( 0,0).

for C1 I did like this x= 2 and y= 1+3t and I integrated it in the given function. I got stuck to make parametric equation from points (2, 4) to points (0, 0) . it is going on decreasing direction. can anyone help me in this problem.

junhoma

 

[galactus]

Please post the entire line integral problem. That will most likely better help us see what you need.

But, if a line segment is going from \(\displaystyle (x_{0},y_{0})\) to \(\displaystyle (x_{1},y_{1})\)

we have \(\displaystyle x=(1-t)x_{0}+tx_{1}\)

\(\displaystyle y=(1-t)y_{0}+ty_{1}\)

\(\displaystyle 0\leq t\leq 1\)

So, if we go from (2,4) to (0,0), the line equations are

\(\displaystyle x=(1-t)\cdot 2+t(0)=2-2t\)

\(\displaystyle y=(1-t)\cdot 4+t(0)=4-4t\)

and it is heading toward the origin.

If it is going from (0,0) to (2,4), then we have

\(\displaystyle x=(1-t)(0)+t(2)=2t\)

\(\displaystyle y=(1-t)(0)+t(4)=4t\)

It is heading away from the origin.

 

[junhoma]

I have to integrate C: x^2ydx+3xdy over C1: points(2,1) to (2,4) and C2 points(2,4) to (0,0)
for c1 I got 18 but I was stuck for C2. I got some idea from your reply.

 

[galactus]

Okey-doke.

\(\displaystyle \int_{C}(x^{2}ydx+3xdy)\)

\(\displaystyle C_{1}: \;\ (0,0)\rightarrow (2,1)\)

\(\displaystyle x=2t, \;\ y=t\)

\(\displaystyle C_{2}: \;\ (2,1)\rightarrow (2,4)\)

\(\displaystyle x=2, \;\ y=1+3t\)

\(\displaystyle C_{3}: \;\ (2,4)\rightarrow (0,0)\)

\(\displaystyle x=2-2t, \;\ y=4-4t\)

\(\displaystyle 0\leq t\leq 1\)

\(\displaystyle C_{1}=\int_{0}^{1}(2t)^{2}(t)dt+3\int_{0}^{1}(2t)(1)dt=5\)

\(\displaystyle C_{2}=\int_{0}^{1}(1+3t)(0)dt+3\int_{0}^{1}(2)(3)dt=18\)

\(\displaystyle C_{3}=\int_{0}^{1}(2-2t)^{2}(4-4t)(-2)dt+3\int_{0}^{1}(2-2t)(-4)dt=-20\)

Add them all up and get \(\displaystyle \boxed{5+18-20=3}\)

Does that set up jive with yours?. It's easy to make a mistake in all this. Remember your dx's and dy's for each C line segment.

 

[junhoma]

Thank you sir. But for C3 I got difference answer . When I did from (2,4) to (0,0) i got negative 44/3 but for same segment when i did from (0,0) to (2,4) I got 16 . I think for same segment the answer should be same either we go positive or reverse direction. I am waiting your respond.

 

[galactus]

Assuming my path is correct and C1, C2, C3 are how I have drawn in the graph.

C1=5, C2=18, C3=-20

Total = 3

Going the opposite direction from (0,0) to (2,4) the solution is 20 instead of -20. As it is expected to be.

Going the opposite direction, the signs should be opposite of what they are.

 

[junhoma]

Your path is correct. But for C3 i got -negative 44/3

 

[galactus]

Check out my C3 and see what you have different.

It should be -20. I just checked it with my line integral program.

 

[junhoma]

Thank you sir,

I got same answer negative 20 like you.
I have one conceptual question with you.
Suppose i have to evaluate x^2y ds over the path C : C is arc of x^2+y^2 =4 and moves clockwise direction from (-2,0) to ( 2, 0)

In this situation can I write x= 2cost , y= 2 sint and t goes from pi to 0. i know if it is a counterclockwise direction t goes from 0 to pi and I can write x= 2cost and y= 2sint. can you please give me some insight how to write parametric equation same path if it goes clockwise and counterclockwise.

please I would be grateful with you.

 

[galactus]

If you evaluated \(\displaystyle \int_{C}x^{2}ydx\) over the circle \(\displaystyle x^{2}+y^{2}=4\).

You could use \(\displaystyle x(t)=2cos(t), \;\ 2sin(t)\) if going counterclockwise.

Use \(\displaystyle x(t)=-2cos(t), \;\ 2sin(t)\) if heading clockwise.

\(\displaystyle ds=\sqrt{(x'(t))^{2}+(y'(t))^{2}}\)

You will get the opposite sign if integrating from \(\displaystyle {\pi}\rightarrow 0\) as opposed to \(\displaystyle 0\rightarrow {\pi}\)

\(\displaystyle \int_{0}^{\pi}[(-2cos(t))^{2}(2sin(t))(2)]dt=\frac{32}{3}\)

\(\displaystyle \int_{\pi}^{0}[(-2cos(t))^{2}(2sin(t))(2)]dt=\frac{-32}{3}\)

 

[junhoma]

Thank you so much. Now I am clear.

How I can get mathematical notation like integral sign and others. Sothat nexttime I could use those sign to make you easy.

 

[galactus]

That is LaTex. You can type the code. To see the code used to make mine display that way, click on 'quote' at the upper right corner of the post.

You can see \int \frac and so forth