I have the following problem on a homework reading and am unsure how to attempt to solve it:
Integrate $f(x,y) = x^2\sin(y)$ over the graph of $g(x,y) = 2x-2y$, on the domain $[0,1]$x$[0,\pi]$.
The book has a sample formula, but I'm not understanding it and its not explaining any more than the following: $$\int\int_S f(x,y,z)dS=\int\int_D \frac{f(x,y,g(x,y)}{\cos\theta}dxdy$$ $$N=-\frac{dg}{dx}i-\frac{dg}{dy}j+k$$ $$\cos\theta=\frac{N\cdot k}{||N||}$$ $$dS=\frac{dxdy}{n \cdot k}$$
Any help would be greatly appreciated!
Edit This is what I think each part is. $$N = -2i+2j+k$$ $$\cos\theta = \frac{-2i+2j+k \cdot k}{||N||} = \frac{1}{3}$$ $$dS = \frac{dxdy}{\frac{-2i+2j+k}{3}\cdot k}= \frac{dxdy}{\frac{1}{3}}=3dxdy$$ Subbing all this in gives: $$\int\int_S f(x,y,z)dS=\int\int_D 3*f(x,y,g(x,y)dxdy$$
I'm unsure of what to do from here to be honest, partially because my original function is only $f(x,y)$ not $f(x,y,z)$ and where the $n \cdot k$ went from subbing in dxdy.
Got it!
$$\int\int_S f(x,y,z)dS=\int\int_D 3*f(x,y,g(x,y)dxdy=\int\int_D 3*x^2\sin{y}dxdy$$ $$=\int3x^2(-\cos{y})dx$$ $$=x^3(-\cos{y})$$ Plugging in the bounds gives that the function is equal to 2.