The question is the following:
Find the surface area given by the revolution of the curve $y = 1 - x$ with $0 \leq x \leq 1$ around the y axis
In order to do that I made the following "substitution": $x = 1 - y$ and $0 \leq y \leq 1$ so I can have an interval for $y$
Then we have:
$$x = f(y)\cos\theta \\ y = y \\ z = f(y)\sin\theta$$
Since $f(y) = 1-y$, then: $$x = (1-y)\cos\theta \\y = y \\ z = (1-y)\sin\theta$$
Cross product between $r_y$ and $r_{\theta}$:
$$||r_y\cdot r_\theta||= (1-y)\sqrt{2}$$
Finally, the integral:
$$\int\int_D||r_y\cdot r_\theta||\text{dA} = \int^{2\pi}_0\int^1_0(1-y)\sqrt{2}\,\,\,dyd\theta=\pi \sqrt{2}$$
Is it okay to perform the "substitution" I did (assuming the interval of $y$ based on the function and the given interval)? I'm also not sure if this is 100% right, can someone check this please?