Evaluating a Surface Integral for S: How Can the Given Formula be Used?

[wifi]

Problem:

Use the fact that $$\int_S \vec{v} \cdot d\vec{S}=\int_S \vec{v} \cdot \frac{\nabla f}{\partial f/\partial x} dy\ dz$$

to evaluate the integral for $S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}$ and $\vec{v}=(3z^2, 6, 6xz)$.

Attempt at a Solution:

I'm having trouble setting up this integral. If I knew what $f$ was, I could easily calculate the gradient, as well as the partial wrt to x. I'd still need to figure out the limits of integration though.

 

[arildno]

f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0

 

[wifi]

f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
f(x,y,z)=0

Hmm. I'm not sure you'd get this from the info given. Any hints?

 

[arildno]

Because
1. It fits. Sort of (I admit I haven't seen closely, though).
2. Otherwise, it would be utterly meaningless, since you would have no way to evaluate the second expression due to lack of knowledge of f.

 

[wifi]

More specifically I meant, how would you find $f$ analytically from this info given?

 

[pasmith]

More specifically I meant, how would you find $f$ analytically from this info given?

From here:

$S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}$

If $y = x^2$ then $y - x^2 = 0$.

(Your inequalities are the wrong way round: you want $0 \leq x \leq 2$ etc.)

 

[wifi]

From here:



If $y = x^2$ then $y - x^2 = 0$.

(Your inequalities are the wrong way round: you want $0 \leq x \leq 2$ etc.)

So $f(x,y,z)=y-x^2$?

 

[LCKurtz]

I would parameterize the surface as $\vec R(x,z)$ and use the formula$$
\iint \vec v\cdot d\vec S =\pm \iint_{x,z}\vec v \cdot \vec R_x\times \vec R_z~dxdz$$where the choice of signs depends on the orientation of the surface which, by the way, you need to specify.