- #1

DryRun

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**Homework Statement**

http://s1.ipicture.ru/uploads/20120119/gW4x6Mxi.jpg

**The attempt at a solution**

I drew the graph. It's a vertical cylinder, with its axis along the z-axis, with the top cut off by the plane z=3 but the cylinder goes down to infinity. The radius of the cylinder is 1.

[tex]\phi(x,y,z)=x^2+y^2-1[/tex]

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

Since the direction of unit normal vector is downward, [itex]-∇\vec{\phi}=-2x\vec{i}-2y\vec{j}[/itex]

[tex]\hat{n}=\frac{-∇\vec{\phi}}{|-∇\vec{\phi}|}[/tex]

[tex]\hat{n}=\frac{-x-y}{\sqrt{x^2+y^2}}[/tex]

[tex]Flux=\int\int\frac{-x^2-y^2}{\sqrt{x^2+y^2}}\,.d \sigma=-\sqrt{x+y}\,.d \sigma[/tex]

Projecting the surface S onto the yz-plane (i cannot project onto the xy-plane as the cylinder starts from z=0 and goes down to infinity):

[tex]x=\sqrt{1-y^2}[/tex]

[tex]x_y=-\frac{y}{\sqrt{1-y^2}}[/tex]

[tex]x_z=0[/tex]

[tex]d\sigma=\frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]

[tex]S.A.=2\int\int \frac{1}{\sqrt{1-y^2}}\,.dzdy[/tex]

The limits are -∞≤z≤3 and -1≤y≤1 but i think i messed up somewhere above, as these limits look suspicious.

When i visualize this thin slice of the surface at the plane z=3, it seems to me that if i project the thin slice onto the yz-plane, the slice would become invisible, as its thickness is zero. Am i right? But since we're dealing with flux here, then this slice does not have a thickness of zero. Correct?

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