Help setting up surface integral, on the surface $2x + 3y + 6z = 60$

Question

I think that $2x+3y +6z=60$ is a plane in 3D, and I need to integrate $f(x,y,z) = 2x/z$ on this plane.

Wouldn't that mean that the integral would be $dx \, dy \, dz$?

None of the options have $dx \, dy \, dz$ though, they are all just $dx \, dz$ or $dy\,dz$.

I think it might be A since that's the only one that has the right function under the integral

Answer 1

We need to evaluate $\int \int_S \frac{2x}{z} dS$

for the surface $2x + 3y + 6z = 60, x \geq 0, y \geq 0, z \geq 6$.

i.e $x = 30 - \frac{3}{2}y - 3z$ ...(i)

$ds = \sqrt{(\frac {\partial x}{\partial y})^2 + (\frac {\partial x}{\partial z})^2 + 1} \, dydz$

$\frac {\partial x}{\partial y} = -\frac{3}{2}, \frac {\partial x}{\partial z} = -3$

$ds = \frac{7}{2} \, dydz$

Using (i), $\int \int_S \frac{2x}{z} dS = \int \int (\frac{210}{z} - \frac{21y}{2z} - 21) dy dz$.

Also, (i), gives bound of $ 0 \leq y \leq 20-2z, 6 \leq z \leq 10$