I need to find the volume under the surface $$z = a^2 - x^2$$ on domain restricted by these surfaces and planes: $$y = 2x$$ $$x+y=a$$ $$z = 0$$ $$y=0$$
I made a drawing but I cannot even infer what the bounds of integration are.
When I drew all these lines (assuming $a=2$) I got this:
What are the bounds of integration in this case?
On
The integration will have to be done in two parts. As can be seen from the graph (Desmos representation:https://www.desmos.com/calculator/mktcqyonrj) a triangular region is formed between y=0, y=2x and y=a-x. Hence the integrals obtained would be: $$\int_0^{a/3} \int_0^{2x} \int_0^{a^2-x^2} \,dz,dy,dx$$
$$\int_{a/3}^a \int_0^{a-x} \int_0^{a^2-x^2} \,dz,dy,dx$$
The sum of these integrals should give you the total volume.
In the graph, if you can imagine it, CFH is the surface we are looking for. We need to calculate the volume beneath this surface and above z=0.
To find the limit we need to make a sketch of the surface and bounds in the planes $z-x$ and $y-x$.
From here it follows that the volume is given by the sum of the following integrals
$$\int_{-a}^{0}\,dx\int_{0}^{-x+a} (a^2-x^2) \,dy $$
$$\int_{0}^{a/3}\,dx\int_{2x}^{-x+a} (a^2-x^2) \,dy $$
$$\int_{a/3}^{a}\,dx\int_{-x+a}^{2x} (a^2-x^2) \,dy $$