Surface Integral of an Inverted Paraboloid with Radius 2 and Height 4

[DryRun]

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.
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}
|∇\vec{\phi}|=\sqrt{4x^2+4y^2+1}
\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}
Flux=\int\int\frac{-2x^2-2y^2-z}{\sqrt{4x^2+4y^2+1}}\,.d\sigma

Projecting on the xy-plane:
z=4-x^2-y^2
z_x=-2x
z_y=-2y
\sigma=\sqrt{4x^2+4y^2+1}\,.dxdy
S.A=\int\int(-2x^2-2y^2-z)\,.dxdy
From z=4-x^2-y^2,
S.A=\int\int(-x^2-y^2-4)\,.dxdy
Transforming to polar coordinates: x=cosθ, y=sinθ, r=1
S.A=\int^{2\pi}_0\int^1_0(-5)\,.rdrd\theta
The answer i get is -5∏, which is wrong. The correct answer is: 9∏/2, so i don't know where i messed up.

 

[LCKurtz]

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.
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}
|∇\vec{\phi}|=\sqrt{4x^2+4y^2+1}
\hat{n}=\frac{∇\vec{\phi}}{|∇\vec{\phi}|}

WOOPS! The orientation of your surface is given to be directed upwards. Look at the $k$ component of your vector. You need the opposite direction. That's why your sign is wrong.

Flux=\int\int\frac{-2x^2-2y^2-z}{\sqrt{4x^2+4y^2+1}}\,.d\sigma

Projecting on the xy-plane:
z=4-x^2-y^2
z_x=-2x
z_y=-2y
\sigma=\sqrt{4x^2+4y^2+1}\,.dxdy
S.A=\int\int(-2x^2-2y^2-z)\,.dxdy
From z=4-x^2-y^2,
S.A=\int\int(-x^2-y^2-4)\,.dxdy
Transforming to polar coordinates: x=cosθ, y=sinθ, r=1
S.A=\int^{2\pi}_0\int^1_0(-5)\,.rdrd\theta
The answer i get is -5∏, which is wrong. The correct answer is: 9∏/2, so i don't know where i messed up.

Remember you are integrating over an area in the xy plane. Areas require two parameters. $x^2+y^2=1$ only on the boundary of your region. The correct parameterization is $x = r\cos\theta,\, y = r\sin\theta,\, dA=rdrd\theta$ and you can't plug in $r=1$ before you integrate.

 

[DryRun]

OK, it looks like i should start from the beginning:
\phi(x,y,z)=4-x^2-y^2-z
∇\vec{\phi}=\frac{\partial \phi}{\partial x}\vec{i}+\frac{\partial \phi}{\partial y}\vec{j}+\frac{\partial \phi}{\partial z}\vec{k}
which gives me:
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}
I believe this part is correct.

Now, since this problem states that the outward unit normal vector be directed upward... I'm not sure, but I'm going to assume this statement changes the expression to:
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}+\vec{k}
Is that correct? Or should i change the sign across the entire expression? Like this:
∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}
I think it makes more sense to change the sign across the entire expression. But i could be wrong again.
\phi(x,y,z)=4-x^2-y^2-z is equivalent to \phi(x,y,z)=-4+x^2+y^2+z. Correct?
And, \phi(x,y,z)=-4+x^2+y^2+z will yield ∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}
But i dare not go any further with this problem, as i have so many uncertainties.

 

[LCKurtz]

OK, it looks like i should start from the beginning:
\phi(x,y,z)=4-x^2-y^2-z
∇\vec{\phi}=\frac{\partial \phi}{\partial x}\vec{i}+\frac{\partial \phi}{\partial y}\vec{j}+\frac{\partial \phi}{\partial z}\vec{k}
which gives me:
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}-\vec{k}
I believe this part is correct.

Now, since this problem states that the outward unit normal vector be directed upward... I'm not sure, but I'm going to assume this statement changes the expression to:
∇\vec{\phi}=-2x\vec{i}-2y\vec{j}+\vec{k}
Is that correct? Or should i change the sign across the entire expression? Like this:
∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}
I think it makes more sense to change the sign across the entire expression.

Of course. You only have two choices, the gradient vector $\nabla \phi$ or its opposite $-\nabla \phi$. You always have to check and choose the one that agrees with the given orientation of the surface.

\phi(x,y,z)=4-x^2-y^2-z is equivalent to \phi(x,y,z)=-4+x^2+y^2+z. Correct?
And, \phi(x,y,z)=-4+x^2+y^2+z will yield ∇\vec{\phi}=2x\vec{i}+2y\vec{j}+\vec{k}
But i dare not go any further with this problem, as i have so many uncertainties.

Be brave. Either way you have the normal facing the correct direction. Work the dot product out and your integrand should come out $x^2+y^2+4$ which you will change to polar coordinates.

 

[DryRun]

\vec{F}.\hat{n}=\frac{2x^2+2y^2+z}{\sqrt{4x^2+4y^2+1}}
S.A=\int\int(2x^2+2y^2+z)\,.dxdy
S.A=\int\int(x^2+y^2+4)\,.dxdy
The points of intersection of the plane z=3 and z=4-x^2-y^2 are: x^2+y^2=1
Projecting the surface S onto the xy plane. Transforming to polar coordinates:
S.A=\int^{2\pi}_0\int^1_0(r^2+4)\,.rdrd\theta=\int^{2\pi}_0\int^1_0r^3+4r\,.drd\theta=\frac{9\pi}{2}

Thanks, LCKurtz!