Using triple integrals to calculate volume of the body $z \geq x^2 + y^2 \quad, \quad z\leq 2-\sqrt{x^2+y^2}$

Question

I'm asked to calculate the volume of the body defined by $$z \geq x^2 + y^2 \quad, \quad z\leq 2-\sqrt{x^2+y^2}$$ My attempt: I find the intersection:

Using the first condition I get: $$\sqrt{x^2 + y^2} = 2-z \iff x^2 + y^2 = (2-z)^2$$ Using the second condition with the first I get: $$z = 4 - 4z + z^2$$ Solving for z nets me $$z_1 = 4$$ $$z_2 = 1$$ I realize $z_2 = 1$ is the only real intersection.

Switching to polar coordinates, I calculate the volume of the bottom half ($0 \leq z \leq 1$):

$$\int^1_0(\int^{2\pi}_0(\int^{\sqrt{z}}_0r\cdot dr)d\theta))dz = \frac{\pi}{2}$$

I calculate the volume of the top half in the same fashion: $$\int^2_1\int_0^{2\pi}\int_0^{2-z} \approx 4.83\pi$$

The resulting volume is approximately $\frac{\pi}{2} + 4.83\pi = 5.33\pi$ However, the answer is supposed to be $\frac{5}{6}\pi$

Now, I've got two questions.

1. Why did I get $z_1$ in the beginning when it's obviously no intersection between the two graphs?

2. Where am I going wrong in calculating the volume? I know there are better ways to calculate the volume, but I want to know why my way doesn't work to get a better grasp of the concept.

Answer 1

When you search for the intersection it is better to do it like this: $z=x^2+y^2$, so $z=2-\sqrt{z}$. If we put $t=\sqrt{z}$, then we get $t^2+t-2=0$, that is $t=1$ or $t=-2$. It is now clear there is only one solution. You have to be careful with squaring things.

$\displaystyle V=\iiint dxdydz=\iint\limits_{proj}(\int_{x^2+y^2}^{2-\sqrt{x^2+y^2}}dz)dxdy=\int_{0}^{1}\int_{0}^{2\pi}(2-r-r^2)rdrd\phi$,

because the projection is the unit circle. So $0\le r\le1$ and $0\le\phi\le2\pi$.

Answer 2

Better to put the $z$ part on the inside.

Let $r=\sqrt{x^2+y^2}$. Then you need $r^2\leq z\leq 2-r$ or $(r+2)(r-1)=r^2+r-2<0$ so you need $r\in[-2,1]$. But you already know $r\geq 0$ so you really need $r\in[0,1]$.

$$\int_{0}^{2\pi}\int_{0}^{1} r\int_{r^2}^{2-r} \,dz\,dr\,d\theta$$

The result is $$2\pi \int_{0}^{1}r(2-r-r^2)\,dr=2\pi\left(1-\frac{1}{3}-\frac{1}{4}\right)=\frac{5\pi}{6}$$