I'd like to find the volume of the following solid.
The solid enclosed by the paraboloid $z=4-x^2-y^2$ and the plane $x+y+z=1$.
Actually original problem is the following (I made upper problem...)
The solid enclosed by the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$ and $z=-5$
In this case,
(the volume of the solid)=$$ \int_{-1}^1 \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \int_{-5}^{1-x-y}dzdydx$$
Now, in case of
The solid enclosed by the paraboloid $z=4-x^2-y^2$ and the plane $x+y+z=1$.
How can I solve this?
Could you give me some hint, please?
$$\begin{align} z & = 4 - x^2 - y^2\\ 1 - x - y & = 4 - x^2 - y^2\\ 0 & = 3 - x^2 + x - y^2 - y\\ 0 & = 3 - x^2 + x - \frac14 + \frac14 - y^2 + y - \frac14 + \frac14\\ 0 & = 3 - (x - \frac12)^2 + \frac14 - (y - \frac12)^2 + \frac14\\ \frac72 & = (x - \frac12)^2 + (y - \frac12)^2 \end{align}$$
So the region of intersection between the plane and paraboloid lies over a circle of radius $\sqrt{\frac72}$ centered at $(\frac12, \frac12)$. Now you can set up the bounds of a 2-d integral.