(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

http://s1.ipicture.ru/uploads/20120120/eAO1JUYk.jpg

The attempt at a solution

[tex]\int\int \vec{F}.\hat{n}\,ds=\int\int\int div\vec{F}\,dV[/tex]

where dV is the element of volume.

[tex]div\vec{F}=3[/tex]

Now, i need to find dV which (i assume) is the hardest part of this problem.

I've drawn the graph in my copybook and it's a paraboloid cut off by the plane z=3, with the outward unit normal vector pointing upward.

OK, i have 2 ideas for calculating the volume of that spherical shape bounded by the plane z=3:

1. I could use triple integral and transform to spherical coordinates. [itex]dxdydz={\rho}^2 \sin\phi .d\rho d\phi d\theta[/itex]

2. Use the flux formula for surface integral to find the surface area of the curved surface and then multiply it by the height along the z-axis which is 4-3=1.

I tried the first idea but i can't figure it out. So, i'm going with my second idea.

[tex]\phi(x,y,z)=z-4+x^2+y^2[/tex]

[tex]∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}[/tex]

[tex]\hat{n}=\frac{2x\vec{i}+2y\vec{j}+\vec{k}}{\sqrt{4x^2+4y^2+1}}[/tex]

[tex]Flux=\int\int \vec{F}.\hat{n}\,.d \sigma=\int\int \frac{2x^2+2y^2+z}{\sqrt{4 x^2+4y^2+1}}\,.d \sigma[/tex]

[tex]d\sigma = \sqrt{1+4x^2+4y^2}\,.dxdy[/tex]

[tex]S.A.=\int\int (2x^2+2y^2+z)\,.dxdy[/tex]

Finding the points of intersection of the plane z=3 and the paraboloid [itex]z=4-x^2-y^2[/itex], gives [itex]x^2+y^2=1[/itex] and hence r=1.

Transforming to polar coordinates:

[tex]S.A.=\int\int (2x^2+2y^2+z)\,.dxdy=\int\int (4+r^2)r\,.drd\theta=\int^{2\pi}_0 \int^1_0 (4r+r^3)\,.drd\theta=\frac{9\pi}{2}[/tex]

The volume becomes: [itex]\frac{9\pi}{2} \times 1 = \frac{9\pi}{2}[/itex]

[tex]\int\int\int div\vec{F}\,dV=\int\int\int 3\,dV=3\times \frac{9\pi}{2}=\frac{27\pi}{2}[/tex]

This is wrong, as (according to my notes) the correct answer is [itex]\frac{3\pi}{2}[/itex]. I have no idea what mistake/s i made.

EDIT: OK, after thinking more about this, i think i know where i made a mistake. I assumed the height of the paraboloid section to be 1 unit (at its peak), but the height is not constant, so i don't know how to find a value for the height. At this point, i'm thinking about going back to my first idea, which is finding the volume via triple integral using spherical coordinates. Any advice?

If i use triple integral with spherical coordinates:

[itex]z=4-x^2-y^2[/itex] becomes [itex]\rho\cos\phi=4-{\rho}^2 {\sin}^2 \phi[/itex]

This becomes: [itex]{\rho}^2 {\cos}^2 \phi +\rho \cos \phi=4+{\rho}^2[/itex]

It's a dilemma as i'm stuck with 2 variables.

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# Homework Help: Evaluate using divergence theorem

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