Thread: Advanced Integrals

Calculate the following line integrals by hand

1.) Note: this integral is a closed curve (so it has the o in the middle of the integral):

int_(C)(2xy – e^(cos(x^2) + 3x)dx + (sin(3y^2 + e^y) + xy^2)dy, where C is the circle of radius 2 centered at the origin oriented counterclockwise.

2.) These below are NOT a closed curve (so DOESN’T have a o in the middle)

int_(C)(F*dr) (note F and r are bold since they are vectors.. also note, * means dot product),
where F(x,y) = <2xy – 3y^2 + 15x^2, x^2 – 6xy + 2>, where C is the polar spiral r = theta for 0 <= theta <= 8Pi

3.) int_(C)(F*dr), where F(x,y) = <xe^(y) + 2xy, cos(y) – x> and C is the portion of the parabola y = x^2 for -1 <= x <= 2 moving from left to right




WORK: Okay, so I tried #2, but I doubt it's this easy. Especially since I think it has to be done using polar coordinates. I'm stuck. This was my thinking:

F(x,y) = <F_(1)(x,y), F_(2)(x,y)>

x = x(t) ... y = y(t)

dr = <dx, dy>

Line integral int_(C)(F*dr), then, is int_(C)[F_(1)(x,y)dx] + int_(C)[F_(2)(x,y)dy]

So, int_(C)[2xy - 3y^2 + 15x^2dx] + int_(C)[x^2 - 6x + 2dy]

= x^2y - 3xy^2 + 5x^3 + x^2y - 6x + 2y, evaluated from whatever the limits of integration are (C).

Can simplify:

= 5x^3 + 2x^2y - 3xy^2 - 6xy + 2y, again evaluated from whatever limits of integration are.

Yeah..

Originally Posted by fifthrapiers

Calculate the following line integrals by hand

1.) Note: this integral is a closed curve (so it has the o in the middle of the integral):

int_(C)(2xy – e^(cos(x^2) + 3x)dx + (sin(3y^2 + e^y) + xy^2)dy, where C is the circle of radius 2 centered at the origin oriented counterclockwise.

This one is easy. Though it looks hard.
This vector field has a scalar potential.
To see this you can check that curlF=0.

Thus, (abusting terminology) Cauchy Closed Curve theorem says that its line integral is zero.

Originally Posted by ThePerfectHacker

This one is easy. Though it looks hard.
This vector field has a scalar potential.
To see this you can check that curlF=0.

Thus, (abusting terminology) Cauchy Closed Curve theorem says that its line integral is zero.

We haven't learned about curl. Since we have to do these by hand, they can't be "too" complicated. So you're saying if I do this integral out it will equal 0? If you have time, mind showing me?

Also, the other 2 involve potential functions?

Yay, I (almost) I figured it out.

Green's thm says that if we have a simply connected region, R with a boundary C (smooth), and which is positively oriented, then we have M and N (that is, if they have continuous partial derivatives) where they are in an open region containing R...then,

int(Mdx + Ndy), with C as the limit of integration = int(int(N_x - M_y))dydx ..

M = 2xy - e^(cos(x^2)+3x) ; N = sin(3y^2 + e^y) + xy^2

N_x = y^2 ; M_y = 2x

So, int(y^3 - 2xy dx), but I need limits of integration now .. so it should be zero is what you're saying?

Originally Posted by fifthrapiers

Yay, I (almost) I figured it out.

Green's thm says that if we have a simply connected region, R with a boundary C (smooth), and which is positively oriented, then we have M and N (that is, if they have continuous partial derivatives) where they are in an open region containing R...then,

int(Mdx + Ndy), with C as the limit of integration = int(int(N_x - M_y))dydx ..

M = 2xy - e^(cos(x^2)+3x) ; N = sin(3y^2 + e^y) + xy^2

N_x = y^2 ; M_y = 2x

So, int(y^3 - 2xy dx), but I need limits of integration now .. so it should be zero is what you're saying?

Okay, so I got -sqrt(4-x^2)..sqrt(4-x^2) for my limits of integration for dy.. and then for dx from -2..2, and I had Maple do it. Maple got 4Pi, not 0 as an answer.

Originally Posted by fifthrapiers

N_x = y^2 ; M_y = 2x

So, int(y^3 - 2xy dx), but I need limits of integration now .. so it should be zero is what you're saying?

Maybe. It is possible I made a mistake,
No, it should be: iint(y^2 - 2x) dA

The limits of integration is over the region which is a circle at the origin with radius 2.

If you express what I have in polar coordinates you should get,

INT(theta=0 , theta = 2*pi) INT(r=0, r=2) [r^2*sin^2(theta) - 2r*cos(theta)]rdr dtheta

Originally Posted by fifthrapiers

2.) These below are NOT a closed curve (so DOESN’T have a o in the middle)

int_(C)(F*dr) (note F and r are bold since they are vectors.. also note, * means dot product),
where F(x,y) = <2xy – 3y^2 + 15x^2, x^2 – 6xy + 2>, where C is the polar spiral r = theta for 0 <= theta <= 8Pi

Use the Fundamental Theorem of Line Integrals.

Thanks! And yes I got the same answer as you (4Pi) by using your awesome polar change which worked well too.

If you get time can you verify if you got a really high number for #2 and then for #3 I got 58ish

Originally Posted by fifthrapiers

Thanks! And yes I got the same answer as you (4Pi) by using your awesome polar change which worked well too.

If you get time can you verify if you got a really high number for #2 and then for #3 I got 58ish

Let me explain #2 how to evaluate that. And for #3 I will post a new solution.

Note, for #2 we can use the fabolous fundamental theorem of line integrals.

We start at theta =0 so r=0. But that means we start at the origin.

We end at theta = 8*pi. Now, x=r*cos theta and y=r*sin theta. So x=8*pi *cos (8pi) = 8*pi and y=8*pi sin(8*pi)=0

So you end at (8pi,0).

1)Evaluate scalar potential at (8pi,0)
2)Evaluate scalar potential at (0,0)
3)Substract #2 from #1.
4)#3 is the answer.

Here is #3. You are a big boy you can do the integrals yourself. I assume.