# Thread: line integral urgent help!

1. ## line integral urgent help!

use line integral to evaluate the integral of D of x^2ydx - y^2xdy

where D is the boundary of the region in the first quadrant , enclosed between the coordinate axes and the circle x^2 + y^2 =16.

Thank you very much.

2. Originally Posted by kittycat
use line integral to evaluate the integral of D of x^2ydx - y^2xdy

where D is the boundary of the region in the first quadrant , enclosed between the coordinate axes and the circle x^2 + y^2 =16.

Thank you very much.
parameterize the curves that form the region. for the circular part, use $x = 4 \cos t$ and $y = 4 \sin t$ for $0 \le t \le \frac {\pi}2$

so, $dx = -4 \sin t$ and $dy = 4 \cos t$

then plug those into your integral and solve.

but you're not done. you need to parameterize the line $y = 0$ for $0 \le x \le 4$ and the line $x = 0$ for $0 \le y \le 4$ and do the respective integrals for those. the required line integral is the sum of these three integrals

can you continue?

i have to run off now, i can't finish

3. Hi Jhevon,

my friend and i worked out different answer for this question. And can't figure out each mistake. You have almost having a fight for that.

Thank you very much.

4. Originally Posted by kittycat
Hi Jhevon,

my friend and i worked out different answer for this question. And can't figure out each mistake. You have almost having a fight for that.

Thank you very much.
don't fight. there are other ways to settle a dispute. in this case, you could verify who has the correct answer using Green's theorem

the answer will be $-32 \pi$

5. HI Jhevon,

You mean like that
=double integral of D of (-y^2 -x^2)dA
=-16 double integral of D dA
=-16* AREA of D
=-16*(pi*4^2)/4
=-64pi

um... i don't get -32pi when using Green's Theorem. Why?

6. Originally Posted by kittycat
HI Jhevon,

You mean like that
=double integral of D of (-y^2 -x^2)dA
=-16 double integral of D dA
=-16* AREA of D
=-16*(pi*4^2)/4
=-64pi

um... i don't get -32pi when using Green's Theorem. Why?
By Green's theorem,

$\int_D x^2y~dx - y^2 x ~dy = - \int_0^{\frac {\pi}2} \int_0^4 r^3 ~dr d \theta = -32 \pi$

Using line integrals:

$\int_D x^2 y~dx - y^2 x~dy = \int_{D_1} x^2 y~dx - y^2 x~dy + \int_{D_2} x^2 y~dx - y^2 x~dy ~+$ $\int_{D_3} x^2 y~dx - y^2 x~dy = 0 - 32 \pi + 0 = -32 \pi$

where $D_1$ is the line $y = 0$ for $0 \le x \le 4$ parameterized by $r(t) = \left< 4t , 0 \right>$ for $0 \le t \le 1$

$D_2$ is the arc of the circle with radius 4 in the first quadrant, parameterized by $r(t) = \left< 4 \cos t, 4 \sin t \right>$ for $0 \le t \le \frac {\pi}2$

and

$D_3$ is the line $x = 0$ for $0 \le y \le 4$ parameterized by $r(t) = \left< 0, 4 - 4t \right>$ for $0 \le t \le 1$

(I assumed we have positive orientation and thus traversed the curve in the anti-clockwise direction)

7. ====================================

======================================HI Jhevon,

=double integral of D of (-y^2 -x^2)dA
=-16 double integral of D dA
=-16* AREA of D
=-16*(pi*4^2)/4
=-64pi

I am also using green's , but I use the fact that x^2 + y^2 =16 to replace x^2 + y^2 by 16 in line 2 . Also replace the double integral of D dA by the area of D ( the are of D should be (pi*4^2)/4 since this is a quarter of the circle) ... um... then I got -64 pi .

I understand your way to Green's and you got -32pi.

Can you see why I am wrong from my way? Please let me know why I am wrong if possible.

Thank you very much.

8. Originally Posted by kittycat
====================================

======================================HI Jhevon,

=double integral of D of (-y^2 -x^2)dA
=-16 double integral of D dA
=-16* AREA of D
=-16*(pi*4^2)/4
=-64pi

I am also using green's , but I use the fact that x^2 + y^2 =16 to replace x^2 + y^2 by 16 in line 2 . Also replace the double integral of D dA by the area of D ( the are of D should be (pi*4^2)/4 since this is a quarter of the circle) ... um... then I got -64 pi .

I understand your way to Green's and you got -32pi.

Can you see why I am wrong from my way? Please let me know why I am wrong if possible.

Thank you very much.
the double integral will not give the area of the region, the single integral does that. that's your problem. if you find the integral the hard way (trig substitution involved) you get my answer

This is my 52th post!!!!!

9. This is my 52th post!!!!!

=======================

WOW WOW WOW

you are an excellent teacher !!!

10. Originally Posted by kittycat
This is my 52th post!!!!!

=======================

WOW WOW WOW

you are an excellent teacher !!!
He thinks so, anyway. But don't tell him I said that.

-Dan

11. Originally Posted by Jhevon
the double integral will not give the area of the region, the single integral does that. that's your problem.
ok, so i'm not sure if what i said here makes too much sense. the double integral of the function 1 is indeed the area of the region we are integrating over. i'm not exactly sure why doing it your way gives twice the answer. i'll have to think about it some more. (the thought popped into my head during my class today, since we were doing similar things in class and my professor used a similar method as you did. but i had to pay attention so i couldn't think about this problem too much)

i'm wondering if it was somehow invalid to change x^2 + y^2 to 16, but i don't see why it should be

Originally Posted by topsquark
He thinks so, anyway. But don't tell him I said that.

-Dan
i heard that

12. the double integral of the function 1 is indeed the area of the region we are integrating over. i'm not exactly sure why doing it your way gives twice the answer. i'll have to think about it some more. (the thought popped into my head during my class today, since we were doing similar things in class and my professor used a similar method as you did. but i had to pay attention so i couldn't think about this problem too much)

Can you figure out why now?
I think my method is fine, but I can't figure out what is the problem by using the double integral = area of D in this question.

============================================

i'm wondering if it was somehow invalid to change x^2 + y^2 to 16, but i don't see why it should be

This part is not the problem. Since I had tried with the other questions, but still return the correct answers.

=======================================
Originally Posted by topsquark
He thinks so, anyway. But don't tell him I said that.

-Dan

i heard that

Dan is joking !

As least to me you are really a great teacher.

I don't like my professor. Whatever he teaches, we don't understand him. Whatever questions we ask him , he only returns one hint and you got to find out yourself or sometimes spending one or two days still don't understand his hint.

Honestly, without you guys(especially you ) in this forum, I don't think I can passes my course!

13. Originally Posted by kittycat
Can you figure out why now?
nope. but i'll keep thinking. i'm going to bed soon. hopefully by tomorrow i can answer you. i'm sure it's something simple we are both overlooking

I think my method is fine, but I can't figure out what is the problem by using the double integral = area of D in this question.
i think so too.

Dan is joking !
i know

As least to me you are really a great teacher.

I don't like my professor. Whatever he teaches, we don't understand him. Whatever questions we ask him , he only returns one hint and you got to find out yourself or sometimes spending one or two days still don't understand his hint.

Honestly, without you guys(especially you ) in this forum, I don't think I can passes my course!