Consider the volume integral over the solid $ \ S \ $ given by $ \ \int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{z=\sqrt{x^{2}+y^{2}}}^{1} dzdydx $.
Identify the Solid $ \ S \ $. $$ $$ I thought $ S $ is not a single solid but combination of two surfaces. One is the Cone $ \ z=\sqrt{x^{2}+y^{2}} $ . But what is the other surface ? Any help please ?
Imagine a cone pointing vertex down.
Now imagine a line segment going from the cone at $z=\sqrt{x^2+y^2}$ to $z=1$ at $(x,y)$.
This is represented by the integral $\displaystyle\int_{\sqrt{x^2+y^2}}^{1}\,dz$.
Next, construct a line segment for every $y$ from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$.
You will get a hyperbolic slice (parallel to the x axis), represented by $\displaystyle \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{1} \,dz \,dy$
Now, create a cone slice for every $x $ from $-1$ to $1$.
Finally you get the region above the cone $z^2=x^2+y^2$ and below the plane $z=1$.
This is represented by $\displaystyle \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{1} \,dz \,dy \,dx$.
You are hinted at a cone because of the equation of $z$. Notice that you integrate $f(x,y,z)=1$ from the cone to $1$, suggesting the plane $z=1$. The boundaries of the first two integrals suggest a circle, which is the intersection of the cone and the plane. Therefore, the volume is that between the cone and the plane.
