Find the volume by using triple integral

Question

Calculate the volume of the contents of the bowl $K$, which is given by $x^2 + y^2 \leq z \leq 1$.

$$x^2+y^2\leq z\leq 1$$

$$x^2+y^2=1\iff y=\pm\sqrt{1-x^2}$$

$$-1\leq x\leq 1$$

$$-\sqrt{1-x^2}\leq y\leq\sqrt{1-x^2}$$

$$\iiint 1~\mathrm{d}x~\mathrm{d}y~\mathrm{d}z=\iint\left(\biggl[z\biggr]_{x^2+y^2}^y\right)~\mathrm{d}x~\mathrm{d}y=\iint\left(-x^2+y^2)~\mathrm{d}y\right)~\mathrm{d}x=\int\biggl[y-yx^2+\frac{y^3}{3}\biggr]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}=\int\left(\sqrt{1-x^2}-\sqrt{1-x^2}x^2+\frac{\sqrt{1-x^2}^3}{3}\right)-\left(-\sqrt{1-x^2}+\sqrt{1-x^2}x^2-\frac{\sqrt{1-x^2}^3}{3}\right)=\int_{-1}^12\sqrt{1-x^2}~\mathrm{d}x-\int_{-1}^12\sqrt{1-x^2}x^2+\int_{-1}^1\frac{\sqrt{1-x^2}}{3}$$

I don't know if I did right so far but as you can see I got something that is kinda hard to integrate and I'm pretty sure I wasn't meant to solve it this way so can someone please explain?

Answer 1

Since:

$$ x^2+y^2 \le z \le 1 \quad \quad \Leftrightarrow \quad \quad 0 \le z \le 1 \, \land \, x^2+y^2 \le z $$

it follows that:

$$ V = \int_0^1 \text{d}z \iint\limits_{x^2+y^2 \le z} 1\,\text{d}x\,\text{d}y = \int_0^1 (\pi\,z)\,\text{d}z = \frac{\pi}{2}\,. $$

Answer 2

You can use an substitution \begin{align*} x =& r \cos \phi\\ y =& r \sin \phi \\ z=& z. \end{align*} After that you will get an integral $$\iiint_D \, dxdydz = \int_0^1 \, dr \int_0^{2\pi} d\phi \int_{r^2}^1 r\, dz = 2\pi \int_0^1 (r-r^3)dr =2 \pi \left(\frac{1}{2}-\frac{1}{4}\right)=\frac{\pi}{2}.$$

Answer 3

Alright so we are given the set

$$K=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2\leq z\leq1\},$$

and wish to find its volume $\operatorname{Vol}(K)$. The fact that we are dealing with the quantities $x^2+y^2$ and $z$ suggests a change to cylindrical coordinates might be useful. In particular, we change variables to

$$\begin{cases} x=r\sin\theta,\\ y=r\cos\theta,\\ z=z. \end{cases}$$

Observe then that the defining equation can be rewritten as

$$r^2\leq z\leq 1,$$

and so the bounds of integration will be

$$0\leq r\leq1,\quad 0\leq\theta<2\pi,\quad r^2\leq z\leq1$$

This gives us the volume as the integral

$$\operatorname{Vol}(K)=\int_0^1\int_0^{2\pi}\int_{r^2}^1r~\mathrm{d}z~\mathrm{d}\theta~\mathrm{d}r=2\pi\int_0^1r(1-r^2)~\mathrm{d}r=2\pi\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{\pi}{2},$$

where the factor of $r$ comes from the Jacobian which gets introduced by the change of variables.