Divergence Theorem for Surface Integrals

[bugatti79]

Homework Statement

Folks, have I set these up correctly? THanks
Use divergence theorem to calculate the surface integral \int \int F.dS for each of the following

Homework Equations

\int \int F.dS=\int \int \int div(F)dV

The Attempt at a Solution

a) F(x,y,z)=xye^z i +xy^2z^3 j- ye^z k and sigma is the surface of the box that is bounded by the coordinate planes and planes x=3, y=2 and z=1

Attempt

\int_0^3 \int_0^2 \int_0^1 2xyz^3 dzdydx where div (F)=ye^z+2xyz^3-ye^z


b) F(x,y,z)=3xy^2 i+xe^zj+z^3k and sigma is surface bounded by cylinder y^2+z^2=1 and x=-1 and x=2

Attempt

\int_0^{2\pi} \int_0^{1} \int_{-1}^{2} (3r^2) r dxdrd\theta where div(F)=3y^2+3z^2

c)F(x,y,z)=(x^3+y^3)i+(y^3+z^3)j+(x^3+z^3)k and sigma is sphere of r=2 and centre 0,0

Attempt

\int_0^{2\pi} \int_0^{\pi} \int_0^{2} 3p^4 sin(\phi) dp d\phi d\theta where div(F)=3(x^2+y^2+z^2)...?

Thanks

 

[tiny-tim]

hi bugatti79!

yes, they all look fine ​

(was anything worrying you about that? )

 

[bugatti79]

hi bugatti79!

yes, they all look fine ​

(was anything worrying you about that? )

Thanks. I just wanted to make sure before I proceeded. Here are 3 more trickier ones.

d) $F(x,y,z)=x^2 sin(y) i+x cos (y) j-xz sin(y) k$ and sigma is the surface x^8+y^8+z^8=8
Dont know how to tackle sigma. I thought I could manipulate it using the equation of a sphere.



e) $F(x,y,z)=x^4 i+x^3z^2 j +4y^2z k$ and sigma is the surface of the solid that is bounded by the cylinder x^2+y^2=1 and the planes z=x+2 and z=0

$\displaystyle \int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4x^3+4y^3)r dx dr d\theta=$

$\displaystyle\int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4r^3cos^3(\theta)+4r^2sin^2(\theta))r dx dr d\theta$

where $x=rcos(\theta)$, $y=rsin(\theta)$ and $div(F)=4x^3+4y^2$


f)$F(x,y,z)=\frac{r}{||r||}$ where $r=x i+yj+zk$ and sigma consists of the hemisphere $z=\sqrt{1-x^2-y^2}$ and the disc and $x^2+y^2\le1$ in the xy plane

This looks like its going to be tedious...? $\displaystyle div(F)=\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{1/2}}$...?

Comments on above?

 

[tiny-tim]

hi bugatti79!

Thanks. I just wanted to make sure before I proceeded. Here are 3 more trickier ones.

d) $F(x,y,z)=x^2 sin(y) i+x cos (y) j-xz sin(y) k$ and sigma is the surface x^8+y^8+z^8=8
Dont know how to tackle sigma. I thought I could manipulate it using the equation of a sphere.

erm …

first, what's divF ? ​

e) …

(i assume you meant 4x³ + 4y²zk and dz ?)

yes, that looks ok

f) This looks like its going to be tedious...? $\displaystyle div(F)=\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{1/2}}+\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{1/2}}$...?

just keep going, it's quite easy and it'll simplify out in the end

 

[bugatti79]

Thanks. I just wanted to make sure before I proceeded. Here are 3 more trickier ones.

d) $F(x,y,z)=x^2 sin(y) i+x cos (y) j-xz sin(y) k$ and sigma is the surface x^8+y^8+z^8=8
Dont know how to tackle sigma. I thought I could manipulate it using the equation of a sphere.

hi bugatti79!


erm …

first, what's divF ? ​

$div(F)=0$! What does that mean physically and mathematically?

e) $F(x,y,z)=x^4 i+x^3z^2 j +4y^2z k$ and sigma is the surface of the solid that is bounded by the cylinder x^2+y^2=1 and the planes z=x+2 and z=0

$\displaystyle \int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4x^3+4y^3)r dx dr d\theta=$

$\displaystyle\int_0^{2\pi} \int_0^1 \int_0^{rcos(\theta)+2} (4r^3cos^3(\theta)+4r^2sin^2(\theta))r dx dr d\theta$

where $x=rcos(\theta)$, $y=rsin(\theta)$ and $div(F)=4x^3+4y^2$


Comments on above?

(i assume you meant 4x³ + 4y²zk and dz ?)

Not sure I follow..but should it be

$\int_0^{2\pi} \int_0^{1} \int_0^{r \cos \theta +2} (4x^3+4y^3)dzdA$ where $da=r dr d\theta$..?

Thanks

 

[tiny-tim]

$div(F)=0$! What does that mean physically and mathematically?

mathematically, it means the integral will be 0

physically, it means F is a conserved flow

Not sure I follow..but should it be

$\int_0^{2\pi} \int_0^{1} \int_0^{r \cos \theta +2} (4x^3+4y^3)dzdA$ where $da=r dr d\theta$..?

my "zk" was a copy-and-paste error, it should have stopped at 4x³ + 4y²

and yes the dx (in dA) should have been dz

 

[bugatti79]

mathematically, it means the integral will be 0

physically, it means F is a conserved flow


my "zk" was a copy-and-paste error, it should have stopped at 4x³ + 4y²

and yes the dx (in dA) should have been dz

Thank you sir :-)

 

[HallsofIvy]

Homework Statement

Folks, have I set these up correctly? THanks
Use divergence theorem to calculate the surface integral \int \int F.dS for each of the following

Homework Equations

\int \int F.dS=\int \int \int div(F)dV

The Attempt at a Solution

a) F(x,y,z)=xye^z i +xy^2z^3 j- ye^z k and sigma is the surface of the box that is bounded by the coordinate planes and planes x=3, y=2 and z=1

Attempt

\int_0^3 \int_0^2 \int_0^1 2xyz^3 dzdydx where div (F)=ye^z+2xyz^3-ye^z

You are aware that this is just \nabla F= 2xyz^3 aren't you?

Your integral is just
2\left(\int_0^3 x dx\right)\left(\int_0^2 y dy\right)\left(\int_0^1 z^3 dz\right)

b) F(x,y,z)=3xy^2 i+xe^zj+z^3k and sigma is surface bounded by cylinder y^2+z^2=1 and x=-1 and x=2

Attempt

\int_0^{2\pi} \int_0^{1} \int_{-1}^{2} (3r^2) r dxdrd\theta where div(F)=3y^2+3z^2

c)F(x,y,z)=(x^3+y^3)i+(y^3+z^3)j+(x^3+z^3)k and sigma is sphere of r=2 and centre 0,0

Attempt

\int_0^{2\pi} \int_0^{\pi} \int_0^{2} 3p^4 sin(\phi) dp d\phi d\theta where div(F)=3(x^2+y^2+z^2)...?

Thanks