Calculate the integral $\iiint\limits_{D}dxdydz$ over domain D

Question

Let $a\in (-1,1)$ and domain $D=\{(x,y,z)\in \Bbb R:x^2+y^2+z^2<1,z>a\}$.

I try to calculate the integral $$\iiint\limits_{D}dxdydz$$ with the use of cylindrical coordinates, but I couldn't find the limits of the integration. Do you have any idea about the limits of the integral ?

Answer 1

As in our region $D$ we have $$ x^2+y^2+z^2 < 1 $$ and $$ z > a, $$ so we can take $r$, $\theta$, and $z$ as follows: $$ 0 \leq r < 1, $$ $$ 0 \leq \theta \leq 2 \pi, $$ and $$ a < z < 1. $$ Moreover, we note that, since \begin{align} x &= r \cos \theta, \\ y &= r \sin \theta, \\ z &= z, \end{align} therefore we find that \begin{align} \frac{\partial(x, y, z) }{\partial(r, \theta, z)} &= \left| \begin{matrix} \frac{\partial x}{\partial r } & \frac{\partial x}{\partial \theta } & \frac{\partial x}{\partial z } \\ \frac{\partial y}{\partial r } & \frac{\partial y}{\partial \theta } & \frac{\partial y}{\partial z } \\ \frac{\partial z}{\partial r } & \frac{\partial z}{\partial \theta } & \frac{\partial z}{\partial z } \end{matrix} \right| \\ &= \left| \begin{matrix} \cos \theta & - r\sin \theta & 0 \\ \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{matrix} \right| \\ &= \left| \begin{matrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{matrix} \right| \\ &= r \cos^2 \theta - \left(-r \sin^2 \theta \right) \\ &= r, \end{align} and so \begin{align} \int\int\int_{D}\, dx \, dy\, dz &= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0 }^{r = 1} r d r\, d \theta \, dz \\ &= \int_{z = a}^{z = 1} \int_{\theta = 0}^{\theta = 2 \pi} \frac{1}{2} d \theta \, dz \\ &= \int_{z = a}^{z = 1} \frac{1}{2} (2 \pi ) dz \\ &= \int_{z = a}^{z = 1} \pi dz \\ &= \pi (1 - a). \end{align}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{D \equiv \braces{\pars{x,y,z} \in \mathbb{R}: x^{2} + y^{2} + z^{2} < 1,\ z > a\ \mbox{where}\ a \in \pars{-1,1}}}$.

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\begin{align} &\bbox[5px,#ffd]{\iiint_{D}\dd x\,\dd y\,\dd z} \\[5mm] = &\ \left.\iiint_{\large\mathbb{R}^{3}}\bracks{x^{2} + y^{2} + z^{2} < 1} \bracks{z > a}\dd x\,\dd y\,\dd z \,\right\vert_{\ a\ \in\ \pars{-1,1}} \\[5mm] = &\ \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \bracks{r\cos\pars{\theta} > a}r^{2}\sin\pars{\theta} \,\dd r\,\dd\theta\,\dd\phi \\[5mm] = &\ 2\pi\int_{0}^{1}r^{2}\int_{-1}^{1}\bracks{r\xi > a}\,\dd\xi\,\dd r = 2\pi\int_{0}^{1}r^{2}\int_{-1}^{1}\bracks{\xi > {a \over r}} \,\dd\xi\,\dd r \\[5mm] = &\ 2\pi\int_{0}^{1}r^{2}\braces{% \bracks{{a \over r} < -1} \int_{-1}^{1}\,\dd\xi + \bracks{-1 < {a \over r} < 1} \int_{a/r}^{1}\,\dd\xi}\,\dd r \\[5mm] = &\ 2\pi\int_{0}^{1}r^{2}\braces{% \bracks{r < -a}2 + \bracks{r > \verts{a}}\pars{1 - {a \over r}}}\,\dd r \\[5mm] = &\ 2\pi\verts{a < 0}\int_{0}^{-a}2r^{2} + 2\pi\int_{\verts{a}}^{1}\pars{r^{2} - ar}\,\dd r \\[5mm] = &\ 2\pi\braces{-\bracks{a < 0}{2 \over 3}\,a^{3} + {1 \over 3} - {a \over 2} + {1 \over 2}a^{3} - {1 \over 3}\,\verts{a}^{3}} \\[5mm] = &\ {1 \over 3}\braces{-\bracks{a < 0}4a^{3} + 2 - 3a + 3a^{3} - 2\verts{a}^{3}}\pi \\[5mm] = &\ \bbx{{1 \over 3}\,\pars{a^{3} - 3a + 2}\pi} \\ & \end{align}

Answer 3

This should be your set-up in spherical:

$\int_0^{2\pi}\int_a^1\int_0^{\sqrt{1-z^2}} r\ dr\ dz\ d\theta$

Which integarates as follows:

$\int_0^{2\pi}\int_a^1 \frac 12 (1-z^2) dz\ d\theta\\ \int_0^{2\pi} \frac 12 - \frac 12 a - \frac 16 + \frac 16 a^3\ d\theta\\ \frac {\pi}{3} (2 - 3a + a^3)$

Which can be factored

$\frac {\pi}{3} (1-a)^2(a + 2)$

If we consider $h = 1-a$ as the height of the spherical cap...

$\frac {\pi}{3} (h)^2(3 - h)$

Which matches with the formula.

https://en.wikipedia.org/wiki/Spherical_cap