How can I express f*dA in terms of u and v for the given function?

[feelau]

Homework Statement

Let f(x,y,z)=x^2yz. Express f*dA in terms of u and v.

The Attempt at a Solution

We're given that
x= u+v
y= u-v
z= u^2+v^2.

The main problem I'm having is that I know I need to take partial derivatives in respect to u and v but I'm not sure how. I did an easier problem that required me to find the area element and I had to take the partial of x in respect to u, then the partial of y in respect to u. In that problem the x, y, z were separated but in this case, x y and z are all together so I'm not sure how to separate them and take the partials to find are element. Thanks

 

[Mathgician]

chain rule

 

[feelau]

so do i just substitute all of those equations into x^2yz and take the derivatives once in respect to u and then once in respect to v?

 

[HallsofIvy]

dA is given by the "fundamental vector product": First write the surface as a vector function of the two parameters:
$$\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= (u+v)\vec{i}+ (u-v)\vec{j}+ (u^2+ v^2)\vec{k}$$
Now take the derivatives with respect to each parameter:
$$\vec{r}_u= \vec{i}+ \vec{j}+ 2u\vec{k}$$
and
$$\vec{r}_v= \vec{i}- \vec{j}+ 2v\vec{k}$$

The "fundamental vector product" is the cross product of those two vectors. Since they both are tangent to the circus (in the direction of the "coordinate axes for coordinates u and v) their cross product is perpendicular to the surface and its length is the "differential of surface area"
If I have done the calculation correctly (something I never guarentee!) then
$$dS= 2\sqrt{2u^2+ 2v^2+ 1}dudv$$
and f(u,v)dA is
$$2(u+v)^2(u-v)(u^2+ v^2)\sqrt{2u^2+ 2v^2+ 1}dudv$$

 

[feelau]

hm...interesting, I'll definitely be doing more problems for practice. Thanks