Unsolved Textbook Exercises: Seeking Help and Solutions

[kelvintc]

$${S1}{(x,y,z)}={x^2}{y}+{z-3}$$
$${\nabla}{S1}={2xy}a_{x}+{x^2}a_{y}+a_{z}$$
$$S2(x,y,z)={x}{\log}{z}-{y^2}+{4}$$
$${\nabla}{S2}={\log}{z}a_{x}-{2y}a_{y}+a_{z}\frac{{x}{\log}{e}}{z} ?$$
$${\cos}{\phi}=\frac{{\nabla}{S1}{\cdot}{\nabla}{S2}}{{\mid}{\nabla}{S1}{\cdot}{\nabla}{S2}{\mid}} ?$$

 

[arildno]

Precisely!
Now, you must figure out the unit normals at the intersection point, and calculate the angle between them. (log(e)=1, BTW)

 

[kelvintc]

log e is not in base 10? it's in base e ? i get the answer correct if log e = 1..

 

[arildno]

Sure:
Now, the natural logarithm of a number "a" is a number b=log(a), so that:
$$e^{b}=a$$ (right?)
If a=e, we have:
$$e^{b}=e=e^{1}\to{b}=log(e)=1$$

 

[kelvintc]

ic.. thanks...

 

[kelvintc]

oic... about no.5 it asks me to prove the Stoke's Theorem.
i get $$\nabla\times{F}={-x^2}{a}_{z}$$
to find $$\int(\nabla\times{F})\cdot{dS}$$ , do i need to separate the triangle to 2 pieces, one is x=0 to x=1, y=0 to y=1 and another one is x=1 to x=2, y=1 to y=0 ?
I can't get 7/6 with both ways i did.
But, $$\oint{F}\cdot{dl}$$ i got 7/6.. curious..

 

[kelvintc]

i'm taking electrical engineering...

 

[arildno]

1.We agree on the expression on the curl
2. I don't see the need to split up the triangle (I'll get into that)
3. I took a master's in fluid mechanics some time ago..

 

[arildno]

My choices for the two triangles:
0<=x<=1,0<=y<=x triangle 1

1<=x<=2, 0<=y<=2-x triangle 2

 

[kelvintc]

yet, i can't get the 7/6. my answer for triangle 1 = -1/4, then triangle 2 = -19/40... sums up can't get 7/6.. anything wrong?

master! oh, it's incredible..

 

[kelvintc]

about (1) since z=0, $$a_{r}$$ has no $$a_{x}$$ component , $$a_{\phi} = -\sin{\phi}{a}_{x} + \cos{\phi}{a}_{y}$$
$$e^{-2}\sin\frac{\phi}{2}*-\sin\phi=-0.102$$

 

[kelvintc]

no.6 i translate all x,y,z into cylindrical coordinates.
$$x = r\cos\phi$$
$$y = r\sin\phi$$
$$z = z$$
$$a_{x} = \cos{\phi}{a}_{r} - \sin{\phi}{a}_{\phi}$$
$$a_{y} = \sin{\phi}{a}_{r} + \cos{\phi}{a}_{\phi}$$
$$a_{x} = a_{z}$$
$$dS = rd\phi{d}{r}a_{r}$$
Hence,
$$F_{r} = r^2\cos^3\phi + r^2\sin^3\phi$$
integrate i can't get the answer also..

 

[arildno]

Hi again!
I got 7/6 in 5); I'll take that first.

1. 7/6 is the number you get by the line integral going around the surface in one direction; -7/6 if you go in the opposite direction

2. We let the normal vector in the surface integral be $$\vec{n}=-\vec{a}_{z}$$

3. Hence, we must calculate I:
$$I=\int_{0}^{1}\int_{0}^{x}x^{2}dydx+\int_{1}^{2}\int_{0}^{2-x}x^{2}dydx$$

The first integral is: $$I_{1}=\frac{1}{4}$$
The second integral is: $$I_{2}=\frac{11}{12}$$

This yields 7/6 in total..

 

[kelvintc]

thanks again.. hehe... oh... i really made too much mistakes in calculation.. sad

 

[arildno]

No wonder you're stuck by 6)!

As far as I can see, the surface integral (the flux) on the cylindrical shell is 0.

However, if you consider the fluxes through the top and bottom disks (z=0 and z=2) as well, you'll get for the total flux $$16\pi$$ which seems to agree with the answer..

 

[arildno]

Please post some work on 1)

 

[kelvintc]

replace z=0 and z=2 to get $$(z^2-1)$$ ?
then find $$\int{rd\phi}{dr}$$ ?
by this i get $$16\pi$$...
i also get the surface integral in r direction = 0

 

[kelvintc]

(1) $$a_{r} = \cos{\phi}{a}_{x} + \sin{\phi}{a}_{y}$$
$$a_{\phi} = -\sin{\phi}{a}_{x} + \cos{\phi}{a}_{y}$$
$$a_{z} = a_{z}$$
then z=0, so vector in x direction only depends on phi .
$$-e^{-2}\sin\frac{\phi}{2}\sin\phi=-0.0586$$

 

[arildno]

I'm not sure if your post 47 is a cry for help, or if you got my meaning.
I'll look into 1) later

 

[kelvintc]

i think i got the answer for (6). just not sure whether the method is wrong or not

 

[arildno]

Post in detail how you get the answer in 6)

 

[kelvintc]

i got 4 for z=0 to z=2 for (z^2-1) then integrate get 2pie x r since r=2 then 4 times 4pie get 16pie

 

[arildno]

Ok, you've got two disks you've got to integrate over in 6)

a) The disk placed at z=2.
At that plane, we have :
$$\vec{F}\cdot\vec{n}=(z^{2}-1)|_{(z=2)}=3$$
Or, it's contribution is: $$3*\pi(2)^{2}=12\pi$$

b) The disk placed at z=0
At that plane, we have:
$$\vec{F}\cdot\vec{n}=(1-z^{2})|_{(z=0)}=1$$
Or, it's contribution is: $$\pi(2)^{2}=4\pi$$

Is this how you thought?

 

[kelvintc]

ya... i did it like that

 

[kelvintc]

i had been doing these qs over the past few weeks and there are still some questions remained unsolved.. anyone can help ?
q3, 7, 8, 9, 13, 14, 15