Compute the volume of the bowl.

Question

A bowl is described by by the inequalities

$$z \geq x^2+y^2-1, \quad z\leq \sqrt{x^2+y^2}+1, \quad z\geq 0.$$

Sketch the bowl and compute its volume.

My problem in these types of questions is sketching the area. Is there any good technique/system one can use to figure out how these areas look like?

The RHS of the first intequality can be rewritten as $x^2+y^2=1^2$, so it's a circle with radius $r=1.$ I'm clueless how to figure out the second one.

Answer 1

To sketch the region by hands is always convenient at first considers $x-y$,$y-z$ and $x-z$ planes.

Note that for example that for $y=0$ that is $x-z$ (and similarly for $x=0$) plane the first region is

• $z \geq x^2-1$ that is the region above the parabola $z=x^2-1$

and in x-y plane

• $x^2+y^2 \leq 1$ that is the region inside the circle $x^2+y^2=1$

and so on.

Regarding the second for $y=0$ (and similarly for $x=0$)

• $z\leq \sqrt{x^2}+1=|x|+1, \quad z\geq 0$ that is the region under the cone $z=|x|+1$ and above $x$ axis

and for $z=k>1$

• $k\leq \sqrt{x^2+y^2}+1\implies x^2+y^2\ge(k-1)^2$ is the region outside the circle $x^2+y^2=(k-1)^2$

Answer 2

Hint

The surface that limits the first region is $x^2+y^2=1+z$. For $z=K>-1$ the section of such surface is a circle of radius $r=\sqrt{x^2+y^2}=\sqrt{1+K}$, so the first region is a paraboloid with circular section and vertex at $z=-1$.

The second region is limited by the surface $\sqrt{x^2+y^2}=z-1$ that, for $z=H>1$ is a circle of radius $r=\sqrt{x^2+y^2}=H-1$. So this region is a cone with circular section and vertex at $z=1$.

Given the symmetry around the $z$ axis, the problem become simple using cylindrical coordinates.