How to Determine Integration Bounds for Flux Calculation?

[UrbanXrisis]

Let S be the part of the plane 3x+y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+2j+3k across the surface S.

\int \int F\cdot dS = \int int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA

\int \int \left( -4(-3)-2(-1)+4-3x-y \right) dA
\int \int \left( 18-3x-y \right) dA

how do I find the ends of integration?

 

[siddharth]

UrbanXrisis, I really don't understand what you did.

What I would have done is
\int \int F\cdot dS = \int \int \left(F\cdot \hat{n}\right) \left(\frac{dxdy}{\hat{n}\cdot\hat{k}}\right)

where, \hat{n} is the unit normal to the plane. That is, to evaluate the surface integral, I'm simply projecting the surface to the x-y plane and integrating.

Now to find the ends of integration, just project the surface to the x-y plane. In this case, you will get a triangle. From that, you can find the limits of integration

 

[HallsofIvy]

Let S be the part of the plane 3x+y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+2j+3k across the surface S.

\int \int F\cdot dS = \int int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[\quote]
? There were no "P", "Q", "R" in the problem! If you MEAN 3x+ y+ z= 4 so P= 3, y= 1, z= 1 then say that!

\int \int \left( -4(-3)-2(-1)+4-3x-y \right) dA
\int \int \left( 18-3x-y \right) dA

Again mystifying! Assuming you are doing the integration by projecting down into the xy- plane, then d\sigma^{/rightarrow}= (3i+ j+ k)dxdy so the integrand is (4i+ 2j+ 3k) . (3i+ j+ k) dxdy= 17dxdy<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> how do I find the ends of integration? </div> </div> </blockquote> The limits of integration? Projecting the plane, 3x+y+z=4, into the xy-plane gives 3x+ y+ 0= 4 or y= 4- 3x. When y= 0, x= 4/3 so you would integrate with respect to y from 0 to 4- 3x and then with respect to x from 0 to 4/3.<br /> <br /> However, the integral of &#039;17 dA&#039; is just 17A, or 17 times the area of that triangle. What is the area of a triangle with base 4/3 and height 4?

 

[-EquinoX-]

I have a similar question to this, with F = 7xi+yj+zk and plane z + 4x + 2y = 12. so what I did was (7xi+ yj+ zk) . (4i+ 2j+ k), is this correct?

 

[-EquinoX-]

anyone?