Surface Integrals: Evaluate A.n dS on 2x+y=6 Plane in 1st Octant

[Physgeek64]

Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4

Homework Equations

Integral A.n dS

The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows...

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy

And I'm not sure how to find the limits...

 

[HallsofIvy]

The plane is in the first octant so bounded by x= y= z= 0. Since there is no "z" in the equation of the plane, the plane is parallel to the z- axis so we need another z bound: we are told that the plane is bounded by z= 4. The line 2x+ y= 6 crosses the x-axis at (0, 6) and the y-axis at (3, 0). We can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x. Equivalently, we can take y going from 0 to 6 and, for each y, x= (6- y)/2. In either case, z goes from 0 to 4.

 

[Physgeek64]

The plane is in the first octant so bounded by x= y= z= 0. Since there is no "z" in the equation of the plane, the plane is parallel to the z- axis so we need another z bound: we are told that the plane is bounded by z= 4. The line 2x+ y= 6 crosses the x-axis at (0, 6) and the y-axis at (3, 0). We can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x. Equivalently, we can take y going from 0 to 6 and, for each y, x= (6- y)/2. In either case, z goes from 0 to 4.

Would there not only be two bounds as its a surface integral? Forgive me if I'm being thick...

 

[HallsofIvy]

Yes, the bounds on x and y. I didn't say you had to use the z bounds!

 

[LCKurtz]

I would add that the problem is incompletely stated without the orientation of the surface being given.

 

[LCKurtz]

Upon looking more closely at this problem and proposed solution -- you cannot paramaterize this surface in terms of the variables $x$ and $y$ because they are dependent on each other. I would suggest the parameterization $x = x,~y=6-2x,~z = z$. So the surface is parameterized as $\vec R(x,z) = \langle x, 6-2x, z\rangle$ and use the formula$$
\iint_S \vec A\cdot d\vec S = \pm\int\int \vec A \cdot \vec R_x\times \vec R_z~dxdz$$where the limits are over the $xz$ rectangle and the sign is chosen for the correct orientation (which needs to be specified).

 

[Ekramul Towsif]

pardon me. i faced a similar prblm . could any1 pls xplain what is first octant??

 

[HallsofIvy]

The "first octant" is the octant in an xyz- coordinate system in which all coordinates are positive.

Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4

The "first octant of the plane" doesn't make sense. A plane has four quadrants not octants. I presume that you mean the portion in the first octant of the xyz coordinate, x> 0, y> 0, 2x+ y< 6 and 0< z< 4.

Homework Equations

Integral A.n dS

The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).[/quote]
No, the unit normal is 1/sqrt(5)(2, 1, 0).

A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows...

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy

Since dS includes the factor sqrt(5) (not sqrt(3)) itself, there is really no point in using "n" at all. I would write, rather, $\vec{A}\cdot \vec{dS}$ where $\vec{dS}= (2, 1, 0)dxdy$ so that the integral will be $\int_{x= 0}^3 \int_{y= 0}^{6- 2x} 2y+ 2x dydx$

And I'm not sure how to find the limits...

In the x,y plane, the line 2x+ y= 6 crosses the x-axis at (3, 0) and crosses the y-axis at (0, 6). The gives a triangle with vertices at (0, 0), (3, 0), and (0, 6) in the first quadrant of the xy- plane. You can take x going from 0 to 3 and, for each x, y from 0 to 6- 2x.

 

[Ekramul Towsif]

pardon me. i faced a similar prblm . could any1 pls xplain what is first octant??

thnx

 

[Ray Vickson]

Homework Statement

Evaluate integral A.n dS for A=(y,2x,-z) and S is the surface of the plane 2x+y=6 in the first octant of the plane cut off by z=4

Homework Equations

Integral A.n dS

The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
A.n= 1/sqrt3(2y,2x,0)

So this is where I get stuck. I think it goes as follows...

ds= dxdy

=> double integral 2/sqrt3 y + 2/sqrt3 x dxdy

And I'm not sure how to find the limits...

In your expression $\vec{A} \cdot \vec{n} \, dS$, $dS$ is the surface area-element in the plane $2x+y=6$. Thus $dS = ds \, dz,$ where is $ds = \sqrt{(dx)^2 + (dy)^2}$ is the length-element along the surface perpendicular to the $z$-axis.

 

[LCKurtz]

The "first octant" is the octant in an xyz- coordinate system in which all coordinates are positive.

The "first octant of the plane" doesn't make sense. A plane has four quadrants not octants. I presume that you mean the portion in the first octant of the xyz coordinate, x> 0, y> 0, 2x+ y< 6 and 0< z< 4.

Homework Equations

Integral A.n dS

The Attempt at a Solution

The normal to the plane is (2,1,0) so the unit normal vector is 1/sqrt3 (2,1,0).
No, the unit normal is 1/sqrt(5)(2, 1, 0). Since dS includes the factor sqrt(5) (not sqrt(3)) itself, there is really no point in using "n" at all. I would write, rather, $\vec{A}\cdot \vec{dS}$ where $\vec{dS}= (2, 1, 0)dxdy$ so that the integral will be $\int_{x= 0}^3 \int_{y= 0}^{6- 2x} 2y+ 2x dydx$

This is not correct. As I explained in post #6, this cannot be worked as an xy domain problem.