Evaluate the following surface integral

[wahaj]

Homework Statement

$$\int_S 2z+1 dS$$
where S is the surface
$$z = 16-x^2-y^2 \quad z>0$$

Homework Equations

$$\int_S f dS = \int_S f(S(u,v)) | \frac{\partial S}{\partial u }\times \frac{\partial S}{\partial v } | dudv$$

The Attempt at a Solution

let
$$x= u \quad y=v$$
$$z = 16 - u^2-v^2$$
$$S=(u,v,16-u^2-v^2)$$
$$-4 \le u \le 4 \quad -4\le v \le 4$$
$$\frac{\partial S}{\partial u } = (1, 0, -2u) \quad \frac{\partial S}{\partial v } = (0,1,-2v)$$
$$| \frac{\partial S}{\partial u }\times \frac{\partial S}{\partial v }| = \sqrt(4u^2+4v^2+1)$$
$$\int_{-4}^4 \int_{-4}^4 (2(16-u^2-v^2)+1) \sqrt (4u^2+4v^2+1) dudv$$

using wolfram the answer I get is 2771.99. However this answer is wrong. This is my first time doing surface integrals and parameterizing surfaces so I can't figure out where I am wrong.

 

[LCKurtz]

Homework Statement

$$\int_S 2z+1 dS$$
where S is the surface
$$z = 16-x^2-y^2 \quad z>0$$

Homework Equations

$$\int_S f dS = \int_S f(S(u,v)) | \frac{\partial S}{\partial u }\times \frac{\partial S}{\partial v } | dudv$$

The Attempt at a Solution

let
$$x= u \quad y=v$$

No point in doing that. Just use $x$ and $y$ in the first place.

$$z = 16 - u^2-v^2$$
$$S=(u,v,16-u^2-v^2)$$
$$-4 \le u \le 4 \quad -4\le v \le 4$$

Those limits would describe a square. But the boundary of the surface in the xy plane is $x^2+y^2 = 16$, which isn't a square, whether you call the variables uv or xy. You might consider polar coordinates to evaluate the integral.