Set up a triple integral to compute the volume of the solid in the first octant bounded by the cone $z = 1 - \sqrt{x^2+y^2}$ and the plane $x +y+z=1$.
I wanted to use polar coordinates for this. I got the theta integral to be equal to be zero from $\pi/2$ and the $r$ integral to be $0$ to $1$. But I don’t know how to set up the inner most integral. Any ideas would help.
On
The cone $z = 1 - \sqrt {x^2+y^2}$ has vertex $(0,0,1)$
The plane $x+y+z = 1$ intersects this vertex.
The two surfaces do not define a solid.
Even if they didn't intersect at the vertex, if we used the plane $x+y+z = 0$ it would be "too steep." That is, it would cut the plane at such an angle as to create a half a hyperbola. You would need a plane like $x+y+4z = 0$ to create an a ellipse for the bottom of the cone.
I guess from polar coordinates here you mean cylindrical coordinates.
Level surface of $z = c$ , where $c \in (0,1)$, would have cross-section like this.
$\hskip 2.5 in$
Here ,$\angle ACB = \dfrac{\pi}{4}$. Let $\angle BAC = \theta \Rightarrow \angle ABC = \dfrac{3 \pi}{4} - \theta$, On using sine rule, one can find that $$ x = \dfrac{r}{\sqrt{2}} cosec \bigg( \dfrac{3 \pi}{4} - \theta \bigg) $$ It should be obvious that $ x \le \rho \le r$. It should be obvious that $r = 1 -z$
For the given region $1 - x - y \le z \le 1 - \sqrt{x^2 + y^2} \implies 1 - r(\cos \theta + \sin \theta) \le z \le 1 - r$.
Recall $$dV = \iiint r dr d\theta d z$$ $$ \implies dV = \int^{\pi /2}_0 \int_x^r \int^{1 - r(\cos \theta + \sin \theta}_{1 - r} r dr d \theta d z $$