Solid volume of a cone with triple integrals

Question

The question: the solid in the first octant bounded by the cone $$z=1-{\sqrt{{x^2}+{y^2}}}$$ and the plane $$z+y+x=1$$ Find the volume.

I tried multiple different orders but got different answers. I'd appreciate if you can get me thru the steps.

Answer 1

Hint. Since for $x,y\geq 0$ with $x^2+y^2\leq 1$, $$1-{\sqrt{{x^2}+{y^2}}}\geq 1-x-y,$$ it follows that the solid is given by the cone intersected the first octant minus the tetrahedron of vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Therefore $$\mbox{Volume}=\frac{1}{4}\cdot\frac{\pi}{3}-\frac{1/2}{3}=\frac{\pi-2}{12}.$$ To obtain the same result by using integration, compute $$\int_{x=0}^1\int_{y=0}^{1-x}\int_{z=1-x-y}^{1-{\sqrt{{x^2}+{y^2}}}}dz dy dx+\int_{x=0}^1\int_{y=1-x}^{\sqrt{1-x^2}}\int_{z=0}^{1-{\sqrt{{x^2}+{y^2}}}}dz dy dx\\ =\int_{x=0}^1\int_{y=0}^{\sqrt{1-x^2}}\int_{z=0}^{1-{\sqrt{{x^2}+{y^2}}}}dz dy dx-\int_{x=0}^1\int_{y=0}^{1-x}\int_0^{z=1-x-y}dz dy dx\\ =\int_{\theta=0}^{\pi/2}\int_{\rho=0}^{1}\int_{z=0}^{1-\rho}dz \rho d\rho d\theta-\int_{x=0}^1\int_{y=0}^{1-x}\int_0^{z=1-x-y}dz dy dx\\ =\frac{\pi}{12}-\frac{1}{6}=\frac{\pi-2}{12}$$