Efficient Line Integral Computation on Cartesian Coordinates

[FrogPad]

In my emag course we are reviewing vector calculus. I've forgotton a lot over the summer, so I just want to make sure I'm doing this properly.

question)
\vec E = \hat x y + \hat y x
Evaluate \int \vec E \cdot d\vec l from P_1(2,1,-1) to P_2(8,2,-1) along the parabola x = 2y^2.

sol)
We are in cartesian coordinates, thus:
d\vec l = \hat x dx + \hat y dy
\vec E \cdot d\vec l = ydx + xdy

Our path is:
x=2y^2
y=\sqrt{\frac{x}{2}}

Thus,
\int_2^8 \sqrt{\frac{x}{2}}\,\,dx + \int_1^2 2y^2 \,\,dy = \frac{28}{3}+\frac{14}{3}=14Does everything look ok?

 

[quasar987]

To me yes.

 

[FrogPad]

Cool. Another question if you wouldn't mind checking :)

question)
Evaluate,
\oint_S \hat R 3 \sin \theta \cdot d\vec s

over the surface of a sphere with radius 5 centered at the orgin.

ans)
This question is in spherical coordinates. The notation used is in the format (R, \phi, \theta)

d\vec s = \hat R R^2 \sin \theta \,d\theta d\phi + \hat \phi R \,dR d\theta + \hat \theta R \sin \theta \,dR d\phi

\hat R3\sin \theta \cdot d\vec s = (R^2 \sin \theta d\theta d\phi)(3\sin \theta) = 3R^2 \sin^2 \theta d\theta d\phi

For our surface we have,
R=5
0 \leq \phi \leq 2\pi
0 \leq \theta \leq \pi

The integral becomes,
3(25) \int_0^{2\pi} \int_0^\pi \sin^2 \theta \,\, d \theta d\phi = 75\pi^2

Does this look ok also?

 

[quasar987]

Correct also. Note that a faster approach is to see that since d\vec{s} is the vector ds times the unit vector perpendicular to the surface, it is actually just

\hat{R}ds=\hat{R}R^2sin\theta d\theta d\phi

 

[FrogPad]

Thanks man

I always appreciate your help :)

 

[quasar987]

No prob! It helps me to keep this stuff fresh in my memory too ;)

 

[HallsofIvy]

In my emag course we are reviewing vector calculus. I've forgotton a lot over the summer, so I just want to make sure I'm doing this properly.

question)
\vec E = \hat x y + \hat y x
Evaluate \int \vec E \cdot d\vec l from P_1(2,1,-1) to P_2(8,2,-1) along the parabola x = 2y^2.

sol)
We are in cartesian coordinates, thus:
d\vec l = \hat x dx + \hat y dy
\vec E \cdot d\vec l = ydx + xdy

Our path is:
x=2y^2
y=\sqrt{\frac{x}{2}}

Thus,
\int_2^8 \sqrt{\frac{x}{2}}\,\,dx + \int_1^2 2y^2 \,\,dy = \frac{28}{3}+\frac{14}{3}=14


Does everything look ok?

Yes, but I wouldn't do it that way, because I dislike square roots!
I would do everything in terms of y: \hat l= (2y^2)\hat x+ y\hat y so d\hat l= (4y)dy \hat x+ dy \hat y and \vec E= y \hat x+ x\hat y= y\hat x+ 2y^2 \hat y

\vec E \cdot d\hat l= (4y^2)dy+ (2y^2)dy= 6y^2 dy
Since, to go from (2, 1, -1) to (8, 2, -1), y must go from 1 to 2, the integral is
\int_1^2 6y^2 dy= 2y^3\right|_1^2= 16- 2= 14[/itex]<br /> just as you got.

 

[FrogPad]

Yes, but I wouldn't do it that way, because I dislike square roots!
I would do everything in terms of y: \hat l= (2y^2)\hat x+ y\hat y so d\hat l= (4y)dy \hat x+ dy \hat y and \vec E= y \hat x+ x\hat y= y\hat x+ 2y^2 \hat y

\vec E \cdot d\hat l= (4y^2)dy+ (2y^2)dy= 6y^2 dy
Since, to go from (2, 1, -1) to (8, 2, -1), y must go from 1 to 2, the integral is
\int_1^2 6y^2 dy= 2y^3\right|_1^2= 16- 2= 14[/itex]<br /> just as you got.

<br /> <br /> Nice. I&#039;ll try this on some of the other review questions. Thank you <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />