- #1
DryRun
Gold Member
- 838
- 4
Homework Statement
http://s1.ipicture.ru/uploads/20120118/EHTTIkiQ.jpg
The attempt at a solution
I've drawn the graph in my copybook. It's an inverted paraboloid with radius 2 on the xy-plane height of 4 on the z-axis, which cuts off at plane z=3.
[tex]∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}[/tex]
[tex]|∇\vec{\phi}|=\sqrt{4x^2+4y^2+1}[/tex]
[tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
[tex]Flux=\int\int\frac{-2x^2-2y^2-z}{\sqrt{4x^2+4y^2+1}}\,.d\sigma[/tex]
Projecting on the xy-plane:
[tex]z=4-x^2-y^2[/tex]
[tex]z_x=-2x[/tex]
[tex]z_y=-2y[/tex]
[tex]\sigma=\sqrt{4x^2+4y^2+1}\,.dxdy[/tex]
[tex]S.A=\int\int(-2x^2-2y^2-z)\,.dxdy[/tex]
From [itex]z=4-x^2-y^2[/itex],
[tex]S.A=\int\int(-x^2-y^2-4)\,.dxdy[/tex]
Transforming to polar coordinates: x=cosθ, y=sinθ, r=1
[tex]S.A=\int^{2\pi}_0\int^1_0(-5)\,.rdrd\theta[/tex]
The answer i get is -5∏, which is wrong. The correct answer is: 9∏/2, so i don't know where i messed up.
http://s1.ipicture.ru/uploads/20120118/EHTTIkiQ.jpg
The attempt at a solution
I've drawn the graph in my copybook. It's an inverted paraboloid with radius 2 on the xy-plane height of 4 on the z-axis, which cuts off at plane z=3.
[tex]∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}[/tex]
[tex]|∇\vec{\phi}|=\sqrt{4x^2+4y^2+1}[/tex]
[tex]\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}[/tex]
[tex]Flux=\int\int\frac{-2x^2-2y^2-z}{\sqrt{4x^2+4y^2+1}}\,.d\sigma[/tex]
Projecting on the xy-plane:
[tex]z=4-x^2-y^2[/tex]
[tex]z_x=-2x[/tex]
[tex]z_y=-2y[/tex]
[tex]\sigma=\sqrt{4x^2+4y^2+1}\,.dxdy[/tex]
[tex]S.A=\int\int(-2x^2-2y^2-z)\,.dxdy[/tex]
From [itex]z=4-x^2-y^2[/itex],
[tex]S.A=\int\int(-x^2-y^2-4)\,.dxdy[/tex]
Transforming to polar coordinates: x=cosθ, y=sinθ, r=1
[tex]S.A=\int^{2\pi}_0\int^1_0(-5)\,.rdrd\theta[/tex]
The answer i get is -5∏, which is wrong. The correct answer is: 9∏/2, so i don't know where i messed up.