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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?

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$$\begin{align} z & = 4 - x^2 - y^2\\ 1 - x - y & = 4 - x^2 - y^2\\ 0 & = 3 - x^2 + x - y^2 - y\\ 0 & = 3 - x^2 + x - \frac14 + \frac14 - y^2 + y - \frac14 + \frac14\\ 0 & = 3 - (x - \frac12)^2 + \frac14 - (y - \frac12)^2 + \frac14\\ \frac72 & = (x - \frac12)^2 + (y - \frac12)^2 \end{align}$$

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.

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  • $\begingroup$ Wow.. then the remainder is following, isn't it? $$ $$ Let D is the domain circle you suggested, then the volume is $$\int\int_D{{((4-x^2-y^2)-(1-x-y)})} dA $$ $$ $$ Would you check it for me, please? $\endgroup$ – user143993 Nov 14 '14 at 5:59
  • $\begingroup$ And I have some question concerned with paraboloid. If we cut parabolid with some plane, then the 'cut section' will be a circle always?? $\endgroup$ – user143993 Nov 14 '14 at 6:01
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    $\begingroup$ Your remainder looks good to me. The intersection will be a circle over $x$ and $y$ except when there is no intersection (when the plane is too high) and when the plane is parallel to the z-axis (the intersection will be a parabola, there will be no enclosed volume). $\endgroup$ – NovaDenizen Nov 14 '14 at 6:11
  • $\begingroup$ $$ $$"The intersection will be a circle except..." $$ $$ How can I demonstrate this statement? Is there a geometric(not analytic geometric) method? Could you give me some advice, please? I am REALLY appreciate you. $\endgroup$ – user143993 Nov 14 '14 at 6:29
  • $\begingroup$ When the plane is too high, the combined equation with the completed squares will look like $-C = (x - a)^2 + (y - b)^2$, which has no solution. When the plane is vertical, the plane equation will have no $z$ coefficient, so $z$ can't be used to combine the two equations. $\endgroup$ – NovaDenizen Nov 14 '14 at 6:36

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