I'm just learning double integrals, and all the questions I've done so far have defined $f(x,y)$ like so:
Evaluate $\int\int{(2x-y^2)}dA$ over the region R, where R is the region enclosed by the graphs of $y=-x+1$ ; $y=x+1$ and $y=3$.
However, the next exercise doesn't define $f(x,y)$, it says:
Find the area of the plane figure bounded by the curves $y_1=(x-1)^2$ and $y_2 = 4-(x-3)^2$.
I've sketched it out and have my limits for both integrals, but I don't know what my $f(x,y)$ I'm supposed to be integrating is, so all I have at the moment is this:
$$\int_1^3\int_{(x-1)^2}^{4-(x-3)^2}f(x,y)dydx$$
I know that $f(x,y)$ is the "ceiling" of the shape that I'm trying to integrate, but I don't know how to find it. Any help is appreciated
For finding the area $f(x,y)=1$ and then $$\int_1^3\int_{(x-1)^2}^{4-(x-3)^2}f(x,y)\,dy\,dx = \int_1^3\int_{(x-1)^2}^{4-(x-3)^2}\,dy\,dx$$
This is saying add up all the infinitesimal elements of (cartesian) area $dy\,dx$ (imagine little rectangles) over the required region, which is then the area of that region.