1. ## Line integral problem

I cant seem to get this problem right:

from what i understand, you solve these with respect to x and y by parameterizing the curve which gets me for the first part r(t)=<sqrt(2)t,0> and r'(t)=<sqrt(2),0>. But when i plug it in as (0) and sqrt(2) for y and dx respectively i only get 0 when i evaluate my integral. the correct answer for C1 should be 1/2, What am i doing wrong? also, i havent even tried the 2nd or 3rd line integrals yet, but would the arc from the circle be integrated from 0 to pi? Thanks.

2. Originally Posted by Evan.Kimia
I cant seem to get this problem right:

from what i understand, you solve these with respect to x and y by parametrizing the curve which gets me for the first part r(t)=<sqrt(2)t,0> and r'(t)=<sqrt(2),0>. But when i plug it in as (0) and sqrt(2) for y and dx respectively i only get 0 when i evaluate my integral. the correct answer for C1 should be 1/2, What am i doing wrong? also, i haven't even tried the 2nd or 3rd line integrals yet, but would the arc from the circle be integrated from 0 to pi? Thanks.

I was going to ask if the integral should be $\displaystyle I=\int_C y\ ds$, but I see that your post used an image, so there's likely no typo.

I agree that $\displaystyle \displaystyle \int_{C_1} y\ dx =0$, where $\displaystyle \displaystyle C_1$ goes from $\displaystyle \displaystyle (0,\ 0)$ to $\displaystyle \displaystyle (0,\ \sqrt 2)$ along $\displaystyle \displaystyle x=0$.

3. Originally Posted by SammyS

I was going to ask if the integral should be $\displaystyle I=\int_C y\ ds$, but I see that your post used an image, so there's likely no typo.

I agree that $\displaystyle \displaystyle \int_{C_1} y\ dx =0$, where $\displaystyle \displaystyle C_1$ goes from $\displaystyle \displaystyle (0,\ 0)$ to $\displaystyle \displaystyle (0,\ \sqrt 2)$ along $\displaystyle \displaystyle x=0$.
Dear SammyS,

Along the line $\displaystyle y=0$ from, $\displaystyle x=0$ to $\displaystyle x=\sqrt{2}$ the integration would be zero. But along the line which joins the points (0,0) and (1,1);

$\displaystyle \displaystyle\int_{c}y~dx=\displaystyle\int^{1}_{0 }x~dx=\left[\frac{x^2}{2}\right]^{1}_{0}=\frac{1}{2}$

This may be the result stated.

Dear Evan.Kimia,

When you integrate along the arc, you have to consider limiting points of the arc. In this case the x coordinate varies from, $\displaystyle x=\sqrt{2}~to~x=1$. Hence write the integral interms of x and substitute these boundry points,

$\displaystyle \displaystyle\int_{c}y~dx=\displaystyle\int^{1}_{\ sqrt{2}}\sqrt{2-x^2}~dx$

If you want to convert this into polar coordinates, substitute the relevant transformations. Hope you would be able to continue.

4. Originally Posted by Sudharaka
Dear SammyS,

Along the line $\displaystyle y=0$ from, $\displaystyle x=0$ to $\displaystyle x=\sqrt{2}$ the integration would be zero. But along the line which joins the points (0,0) and (1,1);

$\displaystyle \displaystyle\int_{c}y~dx=\displaystyle\int^{1}_{0 }x~dx=\left[\frac{x^2}{2}\right]^{1}_{0}=\frac{1}{2}$

This may be the result stated.

When you integrate along the arc, you have to consider limiting points of the arc. In this case the x coordinate varies from, $\displaystyle x=\sqrt{2}~to~x=1$. Hence write the integral interms of x and substitute these boundry points,

$\displaystyle \displaystyle\int_{c}y~dx=\displaystyle\int^{1}_{\ sqrt{2}}\sqrt{2-x^2}~dx$

If you want to convert this into polar coordinates, substitute the relevant transformations. Hope you would be able to continue.

Dear Sudharaka,

I was merely addressing Evan.Kimia's comment about the integral over the first part of the path, namely $\displaystyle \displaystyle C_1$.

You make a good point about the overall problem. It should help Evan.Kimia greatly.

5. On the portion of the circle from $\displaystyle (\sqrt{2}, 0)$ to $\displaystyle (1, 1)$, yes, "convert to polar coordinates" as Sudharak suggests. But i would do it immediately in the parameterization: use $\displaystyle x= \sqrt{2}cos(t)$, $\displaystyle y= \sqrt{2}sin(t)$ so that $\displaystyle dx= -\sqrt{2}sin(t)dt$. Of course t goes from 0 to $\displaystyle \pi/4$.

6. If you looked at it from the big picture and did your research, it's known that Green's Theorem can be used to find the area of a region enclosed by a curve C:

$\displaystyle Area = \oint_C-ydx = \oint_Cxdy = \frac{1}{2}\oint_C-ydx + xdy$

So if you if you do the line integral right, your answer should be negative that of the area of the enclosed region. If you do it right the answer for a) is $\displaystyle -\frac{\pi}{4}$